Solution

The given function is $f(x)=5 x-3$

At $x=0, f(0)=5 \times 0-3=3$

$\lim _{x \to 0} f(x)=\lim _{x \to 0}(5 x-3)=5 \times 0-3=-3$

$\therefore \lim _{x \to 0} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$

At $x=-3, f(-3)=5 \times(-3)-3=-18$

$\lim _{x \to-3} f(x)=\lim _{x \to-3}(5 x-3)=5 \times(-3)-3=-18$

$\therefore \lim _{x \to-3} f(x)=f(-3)$

Therefore, $f$ is continuous at $x=-3$

At $x=5, f(x)=f(5)=5 \times 5-3=25-3=22$

$\lim _{x \to 5} f(x)=\lim _{x \to 5}(5 x-3)=5 \times 5-3=22$

$\therefore \lim _{x \to 5} f(x)=f(5)$

Therefore, $f$ is continuous at $x=5$

Solution

The given function is $f(x)=2 x^{2}-1$

At $x=3, f(x)=f(3)=2 \times 3^{2}-1=17$

$\lim _{x \to 3} f(x)=\lim _{x \to 3}(2 x^{2}-1)=2 \times 3^{2}-1=17$

$\therefore \lim _{x \to 3} f(x)=f(3)$

Thus, $f$ is continuous at $x=3$

Solution

(a) The given function is $f(x)=x-5$

It is evident that $f$ is defined at every real number $k$ and its value at $k$ is $k-5$.

It is also observed that, $\lim _{x \to k} f(x)=\lim _{x \to k}(x-5)=k-5=f(k)$

$\therefore \lim _{x \to k} f(x)=f(k)$

Hence, $f$ is continuous at every real number and therefore, it is a continuous function.

(b) The given function is $f(x)=\frac{1}{x-5}, x \neq 5$

For any real number $k \neq 5$, we obtain

$\lim _{x \to k} f(x)=\lim _{x \to k} \frac{1}{x-5}=\frac{1}{k-5}$

Also, $f(k)=\frac{1}{k-5} \quad($ As $k \neq 5)$

$\therefore \lim _{x \to k} f(x)=f(k)$

Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function.

(c) The given function is $f(x)=\frac{x^{2}-25}{x+5}, x \neq-5$

For any real number $c \neq-5$, we obtain

$ \begin{aligned} & \lim _{x \to c} f(x)=\lim _{x \to c} \frac{x^{2}-25}{x+5}=\lim _{x \to c} \frac{(x+5)(x-5)}{x+5}=\lim _{x \to c}(x-5)=(c-5) \\ & \text{ Also, } f(c)=\frac{(c+5)(c-5)}{c+5}=(c-5) \quad(\text{ as } c \neq-5) \\ & \therefore \lim _{x \to c} f(x)=f(c) \end{aligned} $

Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function.

(d) The given function is $f(x)=|x-5|=\begin{cases} 5-x, \text{ if } x<5 \\ x-5, \text{ if } x \geq 5 \end{cases} .$

This function $f$ is defined at all points of the real line.

Let $c$ be a point on a real line. Then, $c<5$ or $c=5$ or $c>5$

Case I: $c<5$

Then, $f(c)=5-c$

$\lim _{x \to c} f(x)=\lim _{x \to c}(5-x)=5-c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all real numbers less than 5 .

Case II : $c=5$

Then, $f(c)=f(5)=(5-5)=0$

$\lim _{x \to 5^{-}} f(x)=\lim _{x \to 5}(5-x)=(5-5)=0$

$\lim _{x \to 5^{+}} f(x)=\lim _{x \to 5}(x-5)=0$

$\therefore \lim _{x \to c^{-}} f(x)=\lim _{x \to c^{+}} f(x)=f(c)$

Therefore, $f$ is continuous at $x=5$

Case III: $c>5$

Then, $f(c)=f(5)=c-5$

$\lim _{x \to c} f(x)=\lim _{x \to c}(x-5)=c-5$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all real numbers greater than 5 .

Hence, $f$ is continuous at every real number and therefore, it is a continuous function.

Solution

The given function is $f(x)=x^{n}$

It is evident that $f$ is defined at all positive integers, $n$, and its value at $n$ is $n^{n}$.

Then, $\lim _{x \to n} f(n)=\lim _{x \to n}(x^{n})=n^{n}$

$\therefore \lim _{x \to n} f(x)=f(n)$

Therefore, $f$ is continuous at $n$, where $n$ is a positive integer.

Solution

The given function $f$ is $f(x)= \begin{cases}x, & \text{ if } x \leq 1 \\ 5, & \text{ if } x>1\end{cases}$

At $x=0$,

It is evident that $f$ is defined at 0 and its value at 0 is 0 .

Then, $\lim _{x \to 0} f(x)=\lim _{x \to 0} x=0$

$\therefore \lim _{x \to 0} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$

At $x=1$,

$f$ is defined at 1 and its value at 1 is 1 .

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}} x=1$

The right hand limit of $f$ at $x=1$ is,

$ \begin{aligned} & \lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(5)=5 \\ & \therefore \lim _{x \to 1^{-}} f(x) \neq \lim _{x \to 1^{+}} f(x) \end{aligned} $

Therefore, $f$ is not continuous at $x=1$

At $x=2$,

$f$ is defined at 2 and its value at 2 is 5 .

Then, $\lim _{x \to 2} f(x)=\lim _{x \to 2}(5)=5$

$\therefore \lim _{x \to 2} f(x)=f(2)$

Therefore, $f$ is continuous at $x=2$

Find all points of discontinuity of $f$, where $f$ is defined by

Solution

$ f(x)=\begin{cases} 2 x+3, \text{ if } x \leq 2 \\ 2 x-3, \text{ if } x>2 \end{cases} . $

It is evident that the given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line. Then, three cases arise.

(i) $c<2$

(ii) $c>2$

(iii) $c=2$

Case (i) $c<2$

Then, $f(c)=2 c+3$

$\lim _{x \to c} f(x)=\lim _{x \to c}(2 x+3)=2 c+3$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<2$

Case (ii) $c>2$

Then, $f(c)=2 c-3$

$\lim _{x \to c} f(x)=\lim _{x \to c}(2 x-3)=2 c-3$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>2$

Case (iii) $c=2$

Then, the left hand limit of $f$ at $x=2$ is,

$\lim _{x \to 2^{-}} f(x)=\lim _{x \to 2^{-}}(2 x+3)=2 \times 2+3=7$

The right hand limit of $f$ at $x=2$ is,

$\lim _{x \to 2^{+}} f(x)=\lim _{x \to 2^{+}}(2 x-3)=2 \times 2-3=1$

It is observed that the left and right hand limit of $f$ at $x=2$ do not coincide.

Therefore, $f$ is not continuous at $x=2$

Hence, $x=2$ is the only point of discontinuity of $f$.

Solution

$ f(x)=\begin{cases} |x|+3=-x+3, \text{ if } x \leq-3 \\ -2 x, \text{ if }-3<x<3 \\ 6 x+2, \text{ if } x \geq 3 \end{cases} . $

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<-3$, then $f(c)=-c+3$

$\lim _{x \to c} f(x)=\lim _{x \to c}(-x+3)=-c+3$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<-3$

Case II:

If $c=-3$, then $f(-3)=-(-3)+3=6$

$\lim _{x \to-3^{-}} f(x)=\lim _{x \to-3^{-}}(-x+3)=-(-3)+3=6$

$\lim _{x \to-3^{+}} f(x)=\lim _{x \to-3^{+}}(-2 x)=-2 \times(-3)=6$

$\therefore \lim _{x \to-3} f(x)=f(-3)$

Therefore, $f$ is continuous at $x=-3$

Case III:

If $-3<c<3$, then $f(c)=-2 c$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(-2 x)=-2 c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous in $(-3,3)$.

Case IV:

If $c=3$, then the left hand limit of $f$ at $x=3$ is,

$\lim _{x \to 3^{-}} f(x)=\lim _{x \to 3^{-}}(-2 x)=-2 \times 3=-6$

The right hand limit of $f$ at $x=3$ is,

$\lim _{x \to 3^{+}} f(x)=\lim _{x \to 3^{+}}(6 x+2)=6 \times 3+2=20$

It is observed that the left and right hand limit of $f$ at $x=3$ do not coincide.

Therefore, $f$ is not continuous at $x=3$

Case V:

If $c>3$, then $f(c)=6 c+2$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(6 x+2)=6 c+2$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>3$

Hence, $x=3$ is the only point of discontinuity of $f$.

Solution

$ f(x)=\begin{matrix} \frac{|x|}{x} \text{ if } x \neq 0 \\ 0, \text{ if } x=0 \end{matrix} . $

It is known that, $x<0 \Rightarrow|x|=-x$ and $x>0 \Rightarrow|x|=x$

Therefore, the given function can be rewritten as

$f(x)=\begin{cases} \frac{|x|}{x}=\frac{-x}{x}=-1 \text{ if } x<0 \\ 0, \text{ if } x=0 \\ \frac{|x|}{x}=\frac{x}{x}=1, \text{ if } x>0 \end{cases} .$

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<0$, then $f(c)=-1$

$\lim _{x \to c} f(x)=\lim _{x \to c}(-1)=-1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x<0$

Case II:

If $c=0$, then the left hand limit of $f$ at $x=0$ is,

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{-}}(-1)=-1$

The right hand limit of $f$ at $x=0$ is,

$\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}}(1)=1$

It is observed that the left and right hand limit of $f$ at $x=0$ do not coincide.

Therefore, $f$ is not continuous at $x=0$

Case III:

If $c>0$, then $f(c)=1$

$\lim _{x \to c} f(x)=\lim _{x \to c}(1)=1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>0$

Hence, $x=0$ is the only point of discontinuity of $f$.

Solution

$ f(x)=\begin{matrix} \frac{x}{|x|}, \text{ if } x<0 \\ -1, \text{ if } x \geq 0 \end{matrix} . $

It is known that, $x<0 \Rightarrow|x|=-x$

Therefore, the given function can be rewritten as

$f(x)=\begin{cases} \frac{x}{|x|}=\frac{x}{-x}=-1, \text{ if } x<0 \\ -1, \text{ if } x \geq 0 \end{cases} .$

$\Rightarrow f(x)=-1$ for all $x \in \mathbf{R}$

Let $c$ be any real number. Then, $\lim _{x \to c} f(x)=\lim _{x \to c}(-1)=-1$

Also, $f(c)=-1=\lim _{x \to c} f(x)$

Therefore, the given function is a continuous function.

Hence, the given function has no point of discontinuity.

Solution

$ f(x)=\begin{matrix} x+1, \text{ if } x \geq 1 \\ x^{2}+1, \text{ if } x<1 \end{matrix} . $

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<1$, then $f(c)=c^{2}+1$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x^{2}+1)=c^{2}+1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<1$

Case II:

If $c=1$, then $f(c)=f(1)=1+1=2$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(x^{2}+1)=1^{2}+1=2$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(x+1)=1+1=2$

$\therefore \lim _{x \to 1} f(x)=f(1)$

Therefore, $f$ is continuous at $x=1$

Case III:

If $c>1$, then $f(c)=c+1$

$\lim _{x \to c} f(x)=\lim _{x \to c}(x+1)=c+1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Hence, the given function $f$ has no point of discontinuity.

Solution

$ f(x)=\begin{matrix} x^{3}-3, \text{ if } x \leq 2 \\ x^{2}+1, \text{ if } x>2 \end{matrix} . $

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<2$, then $f(c)=c^{3}-3$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x^{3}-3)=c^{3}-3$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<2$

Case II:

If $c=2$, then $f(c)=f(2)=2^{3}-3=5$

$\lim _{x \to 2^{-}} f(x)=\lim _{x \to 2^{-}}(x^{3}-3)=2^{3}-3=5$

$\lim _{x \to 2^{+}} f(x)=\lim _{x \to 2^{+}}(x^{2}+1)=2^{2}+1=5$

$\therefore \lim _{x \to 2} f(x)=f(2)$

Therefore, $f$ is continuous at $x=2$

Case III:

If $c>2$, then $f(c)=c^{2}+1$

$\lim _{x \to c} f(x)=\lim _{x \to c}(x^{2}+1)=c^{2}+1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>2$

Thus, the given function $f$ is continuous at every point on the real line.

Hence, $f$ has no point of discontinuity.

Solution

$ f(x)= \begin{cases}x^{10}-1, & \text{ if } x \leq 1 \\ x^{2}, & \text{ if } x>1\end{cases} $

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<1$, then $f(c)=c^{10}-1$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x^{10}-1)=c^{10}-1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<1$

Case II:

If $c=1$, then the left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(x^{10}-1)=1^{10}-1=1-1=0$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(x^{2})=1^{2}=1$

It is observed that the left and right hand limit of $f$ at $x=1$ do not coincide.

Therefore, $f$ is not continuous at $x=1$

Case III:

If $c>1$, then $f(c)=c^{2}$

$\lim _{x \to c} f(x)=\lim _{x \to c}(x^{2})=c^{2}$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Thus, from the above observation, it can be concluded that $x=1$ is the only point of discontinuity of $f$.

Solution

The given function is $f(x)=\begin{cases} x+5, \text{ if } x \leq 1 \\ x-5, \text{ if } x>1 \end{cases} .$

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<1$, then $f(c)=c+5$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x+5)=c+5$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<1$

Case II:

If $c=1$, then $f(1)=1+5=6$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(x+5)=1+5=6$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(x-5)=1-5=-4$

It is observed that the left and right hand limit of $f$ at $x=1$ do not coincide.

Therefore, $f$ is not continuous at $x=1$

Case III:

If $c>1$, then $f(c)=c-5$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x-5)=c-5$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Thus, from the above observation, it can be concluded that $x=1$ is the only point of discontinuity of $f$.

Discuss the continuity of the function $f$, where $f$ is defined by

Solution

The given function is $f(x)=\begin{cases} 3, \text{ if } 0 \leq x \leq 1 \\ 4, \text{ if } 1<x<3 \\ 5, \text{ if } 3 \leq x \leq 10 \end{cases} .$

The given function is defined at all points of the interval $[0,10]$.

Let $c$ be a point in the interval $[0,10]$.

Case I:

If $0 \leq c<1$, then $f(c)=3$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(3)=3$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous in the interval $[0,1)$.

Case II:

If $c=1$, then $f(3)=3$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(3)=3$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(4)=4$

It is observed that the left and right hand limits of $f$ at $x=1$ do not coincide.

Therefore, $f$ is not continuous at $x=1$

Case III:

If $1<c<3$, then $f(c)=4$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(4)=4$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points of the interval $(1,3)$.

Case IV:

If $c=3$, then $f(c)=5$

The left hand limit of $f$ at $x=3$ is,

$\lim _{x \to 3^{-}} f(x)=\lim _{x \to 3^{-}}(4)=4$

The right hand limit of $f$ at $x=3$ is,

$\lim _{x \to 3^{+}} f(x)=\lim _{x \to 3^{+}}(5)=5$

It is observed that the left and right hand limits of $f$ at $x=3$ do not coincide.

Therefore, $f$ is not continuous at $x=3$

Case V:

If $3<c \leq 10$, then $f(c)=5$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(5)=5$

$\lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points of the interval $(3,10]$.

Hence, $f$ is not continuous at $x=1$ and $x=3$

Solution

The given function is $f(x)= \begin{cases}2 x, & \text{ if } x<0 \\ 0, & \text{ if } 0 \leq x \leq 1 \\ 4 x, & \text{ if } x>1\end{cases}$

The given function is defined at all points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<0$, then $f(c)=2 c$

$\lim _{x \to c} f(x)=\lim _{x \to c}(2 x)=2 c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<0$

Case II:

If $c=0$, then $f(c)=f(0)=0$

The left hand limit of $f$ at $x=0$ is,

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{-}}(2 x)=2 \times 0=0$

The right hand limit of $f$ at $x=0$ is,

$\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}}(0)=0$

$\therefore \lim _{x \to 0} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$

Case III:

If $0<c<1$, then $f(x)=0$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(0)=0$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points of the interval $(0,1)$.

Case IV:

If $c=1$, then $f(c)=f(1)=0$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(0)=0$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(4 x)=4 \times 1=4$

It is observed that the left and right hand limits of $f$ at $x=1$ do not coincide.

Therefore, $f$ is not continuous at $x=1$

Case V:

If $c<1$, then $f(c)=4 c$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(4 x)=4 c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Hence, $f$ is not continuous only at $x=1$

Solution

$ f(x)=\begin{matrix} -2, \text{ if } x \leq-1 \\ 2 x, \text{ if }-1<x \leq 1 \\ 2, \text{ if } x>1 \end{matrix} . $

The given function is defined at all points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<-1$, then $f(c)=-2$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(-2)=-2$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<-1$

Case II:

If $c=-1$, then $f(c)=f(-1)=-2$

The left hand limit of $f$ at $x=-1$ is,

$\lim _{x \to-1^{-}} f(x)=\lim _{x \to-1^{-}}(-2)=-2$

The right hand limit of $f$ at $x=-1$ is,

$ \begin{aligned} & \lim _{x \to-1^{+}} f(x)=\lim _{x \to-1^{+}}(2 x)=2 \times(-1)=-2 \\ & \therefore \lim _{x \to-1} f(x)=f(-1) \end{aligned} $

Therefore, $f$ is continuous at $x=-1$

Case III:

If $-1<c<1$, then $f(c)=2 c$

$\lim _{x \to c} f(x)=\lim _{x \to c}(2 x)=2 c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points of the interval $(-1,1)$.

Case IV:

If $c=1$, then $f(c)=f(1)=2 \times 1=2$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(2 x)=2 \times 1=2$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}} 2=2$

$\therefore \lim _{x \to 1} f(x)=f(c)$

Therefore, $f$ is continuous at $x=2$

Case V:

If $c>1$, then $f(c)=2$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(2)=2$

$\lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Thus, from the above observations, it can be concluded that $f$ is continuous at all points of the real line.

Solution

$ f(x)=\begin{matrix} a x+1, \text{ if } x \leq 3 \\ b x+3, \text{ if } x>3 \end{matrix} . $

If $f$ is continuous at $x=3$, then $\lim _{x \to 3^{-}} f(x)=\lim _{x \to 3^{+}} f(x)=f(3)$

Also,

$\lim _{x \to 3^{-}} f(x)=\lim _{x \to 3^{-}}(a x+1)=3 a+1$

$\lim _{x \to 3^{+}} f(x)=\lim _{x \to 3^{+}}(b x+3)=3 b+3$

$f(3)=3 a+1$

Therefore, from (1), we obtain

$3 a+1=3 b+3=3 a+1$

$\Rightarrow 3 a+1=3 b+3$

$\Rightarrow 3 a=3 b+2$

$\Rightarrow a=b+\frac{2}{3}$

Therefore, the required relationship is given by, $a=b+\frac{2}{3}$

Solution

The given function is $f(x)= \begin{cases}\lambda(x^{2}-2 x), & \text{ if } x \leq 0 \\ 4 x+1, & \text{ if } x>0\end{cases}$

If $f$ is continuous at $x=0$, then

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{+}} f(x)=f(0)$

$\Rightarrow \lim _{x \to 0^{-}} \lambda(x^{2}-2 x)=\lim _{x \to 0^{+}}(4 x+1)=\lambda(0^{2}-2 \times 0)$

$\Rightarrow \lambda(0^{2}-2 \times 0)=4 \times 0+1=0$

$\Rightarrow 0=1=0$, which is not possible

Therefore, there is no value of $\lambda$ for which $f$ is continuous at $x=0$

At $x=1$,

$f(1)=4 x+1=4 \times 1+1=5$

$\lim _{x \to 1}(4 x+1)=4 \times 1+1=5$

$\therefore \lim _{x \to 1} f(x)=f(1)$

Therefore, for any values of $\lambda, f$ is continuous at $x=1$

Solution

The given function is $g(x)=x-[x]$

It is evident that $g$ is defined at all integral points.

Let $n$ be an integer.

Then,

$g(n)=n-[n]=n-n=0$

The left hand limit of $f$ at $x=n$ is,

$\lim _{x \to n^{-}} g(x)=\lim _{x \to n^{-}}(x-[x])=\lim _{x \to n^{-}}(x)-\lim _{x \to n^{-}}[x]=n-(n-1)=1$

The right hand limit of $f$ at $x=n$ is,

$\lim _{x \to n^{+}} g(x)=\lim _{x \to n^{+}}(x-[x])=\lim _{x \to n^{+}}(x)-\lim _{x \to n^{+}}[x]=n-n=0$

It is observed that the left and right hand limits of $f$ at $x=n$ do not coincide.

Therefore, $f$ is not continuous at $x=n$

Hence, $g$ is discontinuous at all integral points.

Solution

It is known that if $g$ and $h$ are two continuous functions, then

$g+h, g-h$, and $g . h$ are also continuous.

It has to proved first that $g(x)=\sin x$ and $h(x)=\cos x$ are continuous functions.

Let $g(x)=\sin x$

It is evident that $g(x)=\sin x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x \to c$, then $h \to 0$

$ \begin{aligned} & g(c)=\sin c \\ & \begin{aligned} \lim _{x \to c} g(x) & =\lim _{x \to c} \sin x \\ & =\lim _{h \to 0} \sin (c+h) \\ & =\lim _{h \to 0}[\sin c \cos h+\cos c \sin h] \\ & =\lim _{h \to 0}(\sin c \cos h)+\lim _{h \to 0}(\cos c \sin h) \\ & =\sin c \cos 0+\cos c \sin 0 \\ & =\sin c+0 \\ & =\sin c \end{aligned} \\ & \therefore \lim _{x \to c} g(x)=g(c) \end{aligned} $

Therefore, $g$ is a continuous function.

Let $h(x)=\cos x$

It is evident that $h(x)=\cos x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x \to c$, then $h \to 0$

$h(c)=\cos c$

$ \begin{aligned} \lim _{x \to c} h(x) & =\lim _{x \to c} \cos x \\ & =\lim _{h \to 0} \cos (c+h) \\ & =\lim _{h \to 0}[\cos c \cos h-\sin c \sin h] \\ & =\lim _{h \to 0} \cos c \cos h-\lim _{h \to 0} \sin c \sin h \\ & =\cos c \cos 0-\sin c \sin 0 \\ & =\cos c \times 1-\sin c \times 0 \\ & =\cos c \end{aligned} $

$\therefore \lim _{x \to c} h(x)=h(c)$

Therefore, $h$ is a continuous function.

Therefore, it can be concluded that

(a) $f(x)=g(x)+h(x)=\sin x+\cos x$ is a continuous function

(b) $f(x)=g(x)-h(x)=\sin x-\cos x$ is a continuous function

(c) $f(x)=g(x) \times h(x)=\sin x \times \cos x$ is a continuous function

Solution

It is known that if $g$ and $h$ are two continuous functions, then

(i) $\frac{h(x)}{g(x)}, g(x) \neq 0$ is continuous

(ii) $\frac{1}{g(x)}, g(x) \neq 0$ is continuous

(iii) $\frac{1}{h(x)}, h(x) \neq 0$ is continuous

It has to be proved first that $g(x)=\sin x$ and $h(x)=\cos x$ are continuous functions.

Let $g(x)=\sin x$

It is evident that $g(x)=\sin x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x \to c$, then $h \to 0$

$g(c)=\sin c$

$\lim _{x \to c} g(x)=\lim _{x \to c} \sin x$

$=\lim _{h \to 0} \sin (c+h)$

$=\lim _{h \to 0}[\sin c \cos h+\cos c \sin h]$

$=\lim _{h \to 0}(\sin c \cos h)+\lim _{h \to 0}(\cos c \sin h)$

$=\sin c \cos 0+\cos c \sin 0$

$=\sin c+0$

$=\sin c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is a continuous function.

Let $h(x)=\cos x$

It is evident that $h(x)=\cos x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x \to c$, then $h \to 0$

$h(c)=\cos c$

$ \begin{aligned} \lim _{x \to c} h(x) & =\lim _{x \to c} \cos x \\ & =\lim _{h \to 0} \cos (c+h) \\ & =\lim _{h \to 0}[\cos c \cos h-\sin c \sin h] \\ & =\lim _{h \to 0} \cos c \cos h-\lim _{h \to 0} \sin c \sin h \\ & =\cos c \cos 0-\sin c \sin 0 \\ & =\cos c \times 1-\sin c \times 0 \\ & =\cos c \end{aligned} $

$\therefore \lim _{x \to c} h(x)=h(c)$

Therefore, $h(x)=\cos x$ is continuous function.

It can be concluded that,

$cosec x=\frac{1}{\sin x}, \sin x \neq 0$ is continuous

$\Rightarrow cosec x, x \neq n \pi(n \in Z)$ is continuous

Therefore, cosecant is continuous except at $x=n p, n$ Î $\mathbf{Z}$

$\sec x=\frac{1}{\cos x}, \cos x \neq 0$ is continuous

$\Rightarrow \sec x, x \neq(2 n+1) \frac{\pi}{2}(n \in \mathbf{Z})$ is continuous

Therefore, secant is continuous except at $x=(2 n+1) \frac{\pi}{2}(n \in \mathbf{Z})$

$\cot x=\frac{\cos x}{\sin x}, \sin x \neq 0$ is continuous

$\Rightarrow \cot x, x \neq n \pi(n \in Z)$ is continuous

Therefore, cotangent is continuous except at $x=n p, n \hat{I} \mathbf{Z}$

Solution

$ f(x)=\begin{cases} \frac{\sin x}{x}, \text{ if } x<0 \\ x+1, \text{ if } x \geq 0 \end{cases} . $

It is evident that $f$ is defined at all points of the real line.

Let $c$ be a real number.

Case I:

If $c<0$, then $f(c)=\frac{\sin c}{c}$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(\frac{\sin x}{x})=\frac{\sin c}{c}$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<0$

Case II:

If $c>0$, then $f(c)=c+1$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x+1)=c+1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>0$

Case III:

If $c=0$, then $f(c)=f(0)=0+1=1$

The left hand limit of $f$ at $x=0$ is,

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0} \frac{\sin x}{x}=1$

The right hand limit of $f$ at $x=0$ is,

$ \begin{aligned} & \lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}}(x+1)=1 \\ & \therefore \lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{+}} f(x)=f(0) \end{aligned} $

Therefore, $f$ is continuous at $x=0$

From the above observations, it can be concluded that $f$ is continuous at all points of the real line.

Thus, $f$ has no point of discontinuity.

Solution

$ f(x)= \begin{cases}x^{2} \sin \frac{1}{x}, & \text{ if } x \neq 0 \\ 0, & \text{ if } x=0\end{cases} $

It is evident that $f$ is defined at all points of the real line.

Let $c$ be a real number.

Case I:

If $c \neq 0$, then $f(c)=c^{2} \sin \frac{1}{c}$

$\lim _{x \to c} f(x)=\lim _{x \to c}(x^{2} \sin \frac{1}{x})=(\lim _{x \to c} x^{2})(\lim _{x \to c} \sin \frac{1}{x})=c^{2} \sin \frac{1}{c}$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x \neq 0$

Case II:

If $c=0$, then $f(0)=0$ $\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{-}}(x^{2} \sin \frac{1}{x})=\lim _{x \to 0}(x^{2} \sin \frac{1}{x})$

It is known that, $-1 \leq \sin \frac{1}{x} \leq 1, x \neq 0$

$\Rightarrow-x^{2} \leq \sin \frac{1}{x} \leq x^{2}$

$\Rightarrow \lim _{x \to 0}(-x^{2}) \leq \lim _{x \to 0}(x^{2} \sin \frac{1}{x}) \leq \lim _{x \to 0} x^{2}$

$\Rightarrow 0 \leq \lim _{x \to 0}(x^{2} \sin \frac{1}{x}) \leq 0$

$\Rightarrow \lim _{x \to 0}(x^{2} \sin \frac{1}{x})=0$

$\therefore \lim _{x \to 0^{-}} f(x)=0$

Similarly, $\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}}(x^{2} \sin \frac{1}{x})=\lim _{x \to 0}(x^{2} \sin \frac{1}{x})=0$

$\therefore \lim _{x \to 0^{-}} f(x)=f(0)=\lim _{x \to 0^{+}} f(x)$

Therefore, $f$ is continuous at $x=0$

From the above observations, it can be concluded that $f$ is continuous at every point of the real line.

Thus, $f$ is a continuous function.

Solution

$ f(x)= \begin{cases}\sin x-\cos x, & \text{ if } x \neq 0 \\ -1 & \text{ if } x=0\end{cases} $

It is evident that $f$ is defined at all points of the real line.

Let $c$ be a real number.

Case I:

If $c \neq 0$, then $f(c)=\sin c-\cos c$

$\lim _{x \to c} f(x)=\lim _{x \to c}(\sin x-\cos x)=\sin c-\cos c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x \neq 0$

Case II:

If $c=0$, then $f(0)=-1$

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1$

$\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1$

$\therefore \lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{+}} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$

From the above observations, it can be concluded that $f$ is continuous at every point of the real line.

Thus, $f$ is a continuous function.

Solution

$ f(x)= \begin{cases}\frac{k \cos x}{\pi-2 x}, & \text{ if } x \neq \frac{\pi}{2} \\ 3, & \text{ if } x=\frac{\pi}{2}\end{cases} $

The given function $f$ is continuous at $x=\frac{\pi}{2}$, if $f$ is defined at $x=\frac{\pi}{2}$ and if the value of the $f$ at $x=\frac{\pi}{2}$ equals the limit of $f$ at $x=\frac{\pi}{2}$.

It is evident that $f$ is defined at $x=\frac{\pi}{2}$ and $f(\frac{\pi}{2})=3$

$\lim _{x \to \frac{\pi}{2}} f(x)=\lim _{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}$

Put $x=\frac{\pi}{2}+h$

Then, $x \to \frac{\pi}{2} \Rightarrow h \to 0$

$\begin{aligned} \therefore \lim _{x \to \frac{\pi}{2}} f(x) & =\lim _{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \to 0} \frac{k \cos (\frac{\pi}{2}+h)}{\pi-2(\frac{\pi}{2}+h)} \\ & =k \lim _{h \to 0} \frac{-\sin h}{-2 h}=\frac{k}{2} \lim _{h \to 0} \frac{\sin h}{h}=\frac{k}{2} \cdot 1=\frac{k}{2}\end{aligned}$

$\therefore \lim _{x \to \frac{\pi}{2}} f(x)=f(\frac{\pi}{2})$

$\Rightarrow \frac{k}{2}=3$

$\Rightarrow k=6$

Therefore, the required value of $k$ is 6 .

Solution

The given function is $f(x)= \begin{cases}k x^{2}, & \text{ if } x \leq 2 \\ 3, & \text{ if } x>2\end{cases}$

The given function $f$ is continuous at $x=2$, if $f$ is defined at $x=2$ and if the value of $f$ at $x=2$ equals the limit of $f$ at $x=2$

It is evident that $f$ is defined at $x=2$ and $f(2)=k(2)^{2}=4 k$

$ \begin{aligned} & \lim _{x \to 2^{-}} f(x)=\lim _{x \to 2^{+}} f(x)=f(2) \\ & \Rightarrow \lim _{x \to 2^{-}}(k x^{2})=\lim _{x \to 2^{+}}(3)=4 k \\ & \Rightarrow k \times 2^{2}=3=4 k \\ & \Rightarrow 4 k=3=4 k \\ & \Rightarrow 4 k=3 \\ & \Rightarrow k=\frac{3}{4} \end{aligned} $

Therefore, the required value of $k$ is $\frac{3}{4}$.

Solution

The given function is $f(x)=\begin{cases} k x+1, \text{ if } x \leq \pi \\ \cos x, \text{ if } x>\pi \end{cases} .$

The given function $f$ is continuous at $x=p$, if $f$ is defined at $x=p$ and if the value of $f$ at $x=p$ equals the limit of $f$ at $x=p$

It is evident that $f$ is defined at $x=p$ and $f(\pi)=k \pi+1$

$ \begin{aligned} & \lim _{x \to \pi^{-}} f(x)=\lim _{x \to \pi^{+}} f(x)=f(\pi) \\ & \Rightarrow \lim _{x \to \pi^{-}}(k x+1)=\lim _{x \to \pi^{+}} \cos x=k \pi+1 \\ & \Rightarrow k \pi+1=\cos \pi=k \pi+1 \\ & \Rightarrow k \pi+1=-1=k \pi+1 \\ & \Rightarrow k=-\frac{2}{\pi} \end{aligned} $

Therefore, the required value of $k$ is $-\frac{2}{\pi}$.

Solution

$ f(x)=\begin{cases} k x+1, \text{ if } x \leq 5 \\ 3 x-5, \text{ if } x>5 \end{cases} . $

The given function $f$ is continuous at $x=5$, if $f$ is defined at $x=5$ and if the value of $f$ at $x=5$ equals the limit of $f$ at $x=5$

It is evident that $f$ is defined at $x=5$ and $f(5)=k x+1=5 k+1$

$ \begin{aligned} & \lim _{x \to 5^{-}} f(x)=\lim _{x \to 5^{+}} f(x)=f(5) \\ & \Rightarrow \lim _{x \to 5^{-}}(k x+1)=\lim _{x \to 5^{+}}(3 x-5)=5 k+1 \\ & \Rightarrow 5 k+1=15-5=5 k+1 \\ & \Rightarrow 5 k+1=10 \\ & \Rightarrow 5 k=9 \\ & \Rightarrow k=\frac{9}{5} \end{aligned} $

Therefore, the required value of $k$ is $\frac{9}{5}$.

Solution

$ f(x)= \begin{cases}5, & \text{ if } x \leq 2 \\ a x+b, & \text{ if } 2<x<10 \\ 21, & \text{ if } x \geq 10\end{cases} $

It is evident that the given function $f$ is defined at all points of the real line.

If $f$ is a continuous function, then $f$ is continuous at all real numbers.

In particular, $f$ is continuous at $x=2$ and $x=10$

Since $f$ is continuous at $x=2$, we obtain

$ \begin{aligned} & \lim _{x \to 2^{-}} f(x)=\lim _{x \to 2^{+}} f(x)=f(2) \\ & \Rightarrow \lim _{x \to 2^{-}}(5)=\lim _{x \to 2^{+}}(a x+b)=5 \\ & \Rightarrow 5=2 a+b=5 \\ & \Rightarrow 2 a+b=5 \\ \end{aligned} $

Since $f$ is continuous at $x=10$, we obtain

$$ \begin{align*} & \lim _{x \to 10^{-}} f(x)=\lim _{x \to 10^{+}} f(x)=f(10) \\ & \Rightarrow \lim _{x \to 10^{-}}(a x+b)=\lim _{x \to 10^{+}}(21)=21 \\ & \Rightarrow 10 a+b=21=21 \\ & \Rightarrow 10 a+b=21 \tag{2} \end{align*} $$

On subtracting equation (1) from equation (2), we obtain

$8 a=16$

$\Rightarrow a=2$

By putting $a=2$ in equation (1), we obtain

$2 \times 2+b=5$

$\Rightarrow 4+b=5$ $\Rightarrow b=1$

Therefore, the values of $a$ and $b$ for which $f$ is a continuous function are 2 and 1 respectively.

Solution

The given function is $f(x)=\cos (x^{2})$

This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as,

$f=g \circ h$, where $g(x)=\cos x$ and $h(x)=x^{2}$

$[\because(g \circ h)(x)=g(h(x))=g(x^{2})=\cos (x^{2})=f(x)]$

It has to be first proved that $g(x)=\cos x$ and $h(x)=x^{2}$ are continuous functions.

It is evident that $g$ is defined for every real number.

Let $c$ be a real number.

Then, $g(c)=\cos c$

Put $x=c+h$

If $x \to c$, then $h \to 0$

$\lim _{x \to c} g(x)=\lim _{x \to c} \cos x$

$=\lim _{h \to 0} \cos (c+h)$

$=\lim _{h \to 0}[\cos c \cos h-\sin c \sin h]$

$=\lim _{h \to 0} \cos c \cos h-\lim _{h \to 0} \sin c \sin h$

$=\cos c \cos 0-\sin c \sin 0$

$=\cos c \times 1-\sin c \times 0$

$=\cos c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g(x)=\cos x$ is continuous function. $h(x)=x^{2}$

Clearly, $h$ is defined for every real number.

Let $k$ be a real number, then $h(k)=k^{2}$

$\lim _{x \to k} h(x)=\lim _{x \to k} x^{2}=k^{2}$

$\therefore \lim _{x \to k} h(x)=h(k)$

Therefore, $h$ is a continuous function.

It is known that for real valued functions $g$ and $h$, such that $(g \circ h)$ is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.

Therefore, $f(x)=(g o h)(x)=\cos (x^{2})$ is a continuous function.

Solution

The given function is $f(x)=|\cos x|$

This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as,

$f=g \circ h$, where $g(x)=|x|$ and $h(x)=\cos x$

$[\because(g \circ h)(x)=g(h(x))=g(\cos x)=|\cos x|=f(x)]$

It has to be first proved that $g(x)=|x|$ and $h(x)=\cos x$ are continuous functions.

$g(x)=|x|$ can be written as

$g(x)= \begin{cases}-x, & \text{ if } x<0 \\ x, & \text{ if } x \geq 0\end{cases}$

Clearly, $g$ is defined for all real numbers.

Let $c$ be a real number.

Case I:

If $c<0$, then $g(c)=-c$ and $\lim _{x \to c} g(x)=\lim _{x \to c}(-x)=-c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x<0$

Case II:

If $c>0$, then $g(c)=c$ and $\lim _{x \to c} g(x)=\lim _{x \to c} x=c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x>0$

Case III:

If $c=0$, then $g(c)=g(0)=0$

$ \begin{aligned} & \lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{-}}(-x)=0 \\ & \lim _{x \to 0^{+}} g(x)=\lim _{x \to 0^{+}}(x)=0 \\ & \therefore \lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{+}}(x)=g(0) \end{aligned} $

Therefore, $g$ is continuous at $x=0$

From the above three observations, it can be concluded that $g$ is continuous at all points.

$h(x)=\cos x$

It is evident that $h(x)=\cos x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x \to c$, then $h \to 0$

$h(c)=\cos c$

$ \begin{aligned} \lim _{x \to c} h(x) & =\lim _{x \to c} \cos x \\ & =\lim _{h \to 0} \cos (c+h) \\ & =\lim _{h \to 0}[\cos c \cos h-\sin c \sin h] \\ & =\lim _{h \to 0} \cos c \cos h-\lim _{h \to 0} \sin c \sin h \\ & =\cos c \cos 0-\sin c \sin 0 \\ & =\cos c \times 1-\sin c \times 0 \\ & =\cos c \end{aligned} $

$\therefore \lim _{x \to c} h(x)=h(c)$

Therefore, $h(x)=\cos x$ is a continuous function.

It is known that for real valued functions $g$ and $h$, such that $(g \circ h)$ is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.

Therefore, $f(x)=(g \circ h)(x)=g(h(x))=g(\cos x)=|\cos x|$ is a continuous function.

Solution

Let $f(x)=\sin |x|$

This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as,

$f=g \circ h$, where $g(x)=|x|$ and $h(x)=\sin x$

$[\because(g \circ h)(x)=g(h(x))=g(\sin x)=|\sin x|=f(x)]$

It has to be proved first that $g(x)=|x|$ and $h(x)=\sin x$ are continuous functions.

$g(x)=|x|$ can be written as

$g(x)= \begin{cases}-x, & \text{ if } x<0 \\ x, & \text{ if } x \geq 0\end{cases}$

Clearly, $g$ is defined for all real numbers.

Let $c$ be a real number.

Case I:

If $c<0$, then $g(c)=-c$ and $\lim _{x \to c} g(x)=\lim _{x \to c}(-x)=-c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x<0$

Case II:

If $c>0$, then $g(c)=c$ and $\lim _{x \to c} g(x)=\lim _{x \to c} x=c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x>0$

Case III:

If $c=0$, then $g(c)=g(0)=0$

$ \begin{aligned} & \lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{-}}(-x)=0 \\ & \lim _{x \to 0^{+}} g(x)=\lim _{x \to 0^{+}}(x)=0 \\ & \therefore \lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{+}}(x)=g(0) \end{aligned} $

Therefore, $g$ is continuous at $x=0$

From the above three observations, it can be concluded that $g$ is continuous at all points.

$h(x)=\sin x$

It is evident that $h(x)=\sin x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+k$

If $x \to c$, then $k \to 0$

$h(c)=\sin c$

$h(c)=\sin c$

$ \begin{aligned} \lim _{x \to c} h(x) & =\lim _{x \to c} \sin x \\ & =\lim _{k \to 0} \sin (c+k) \\ & =\lim _{k \to 0}[\sin c \cos k+\cos c \sin k] \\ & =\lim _{k \to 0}(\sin c \cos k)+\lim _{h \to 0}(\cos c \sin k) \\ & =\sin c \cos 0+\cos c \sin 0 \\ & =\sin c+0 \\ & =\sin c \end{aligned} $

$\therefore \lim _{x \to c} h(x)=g(c)$

Therefore, $h$ is a continuous function.

It is known that for real valued functions $g$ and $h$, such that $(g \circ h)$ is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.

Therefore, $f(x)=(g \circ h)(x)=g(h(x))=g(\sin x)=|\sin x|$ is a continuous function.

Solution

The given function is $f(x)=|x|-|x+1|$

The two functions, $g$ and $h$, are defined as

$g(x)=|x|$ and $h(x)=|x+1|$

Then, $f=g-h$

The continuity of $g$ and $h$ is examined first.

$g(x)=|x|$ can be written as

$g(x)= \begin{cases}-x, & \text{ if } x<0 \\ x, & \text{ if } x \geq 0\end{cases}$

Clearly, $g$ is defined for all real numbers.

Let $c$ be a real number.

Case I:

If $c<0$, then $g(c)=-c$ and $\lim _{x \to c} g(x)=\lim _{x \to c}(-x)=-c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x<0$

Case II:

If $c>0$, then $g(c)=c$ and $\lim _{x \to c} g(x)=\lim _{x \to c} x=c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x>0$

Case III:

If $c=0$, then $g(c)=g(0)=0$

$\lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{-}}(-x)=0$

$\lim _{x \to 0^{+}} g(x)=\lim _{x \to 0^{+}}(x)=0$

$\therefore \lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{+}}(x)=g(0)$

Therefore, $g$ is continuous at $x=0$

From the above three observations, it can be concluded that $g$ is continuous at all points.

$h(x)=|x+1|$ can be written as

$h(x)= \begin{cases}-(x+1), & \text{ if, } x<-1 \\ x+1, & \text{ if } x \geq-1\end{cases}$

Clearly, $h$ is defined for every real number.

Let $c$ be a real number.

Case I:

If $c<-1$, then $h(c)=-(c+1)$ and $\lim _{x \to c} h(x)=\lim _{x \to c}[-(x+1)]=-(c+1)$

$\therefore \lim _{x \to c} h(x)=h(c)$

Therefore, $h$ is continuous at all points $x$, such that $x<-1$

Case II:

If $c>-1$, then $h(c)=c+1$ and $\lim _{x \to c} h(x)=\lim _{x \to c}(x+1)=c+1$

$\therefore \lim _{x \to c} h(x)=h(c)$

Therefore, $h$ is continuous at all points $x$, such that $x>-1$

Case III:

If $c=-1$, then $h(c)=h(-1)=-1+1=0$

$\lim _{x \to-1^{-}} h(x)=\lim _{x \to-1^{-}}[-(x+1)]=-(-1+1)=0$

$\lim _{x \to-1^{+}} h(x)=\lim _{x \to-1^{+}}(x+1)=(-1+1)=0$

$\therefore \lim _{x \to-1^{-}} h(x)=\lim _{h \to-1^{+}} h(x)=h(-1)$

Therefore, $h$ is continuous at $x=-1$

From the above three observations, it can be concluded that $h$ is continuous at all points of the real line.

$g$ and $h$ are continuous functions. Therefore, $f=g-h$ is also a continuous function.

Therefore, $f$ has no point of discontinuity.