Let $f$ be a real valued function defined on a domain $D \subseteq \mathbb{R}$

Suppose $a \in D$ (ie. $f(a)$ is defined)

We say that $f$ is continuous at a

if $\quad f(a)=\lim _{x \rightarrow a} f(x)$

Recall that for $\lim _{x \rightarrow a} f(x)$ to exist the function $f(x)$ must be defined in some open interval containing $a$.

(but not necessarily at x=a)

So, $f(x)$ in continuous at $x=a$

(i) if $\lim _{x \rightarrow a} f(x)$ exists

and (ii) $f(a)=\lim _{x \rightarrow a} f(x)$.

So, for $f(x)$ to be continuous at $x=a$

the function must be defined at $x=a$

and $f(a)$ must equal $\lim _{x \rightarrow a} f(x)$.

f(x)= $\begin{cases}0, & x<0 \\ 1, & x \geqslant 0\end{cases}$

Here, $\lim _{x \rightarrow 0} f(x)$ does not

exist (because $\lim _{x \rightarrow 0^{-}} f(x)=0 \neq 1=\lim _{x \rightarrow 0^{+}} f(x)$ )

So, $f(x)$ is not continuous at $x=0$ (even though $f(x)$ is defined at $x=0 ; f(0)=1$ )

But if $a \neq 0$, then $\lim _{x \rightarrow a} f(x)= \begin{cases}1, & \text { if } a>0 \\ 0, & \text { if } a<0\end{cases}$

Also, $f(a)= \begin{cases}1 & \text { if } a>0 \\ 0 & \text { if } a<0\end{cases}$

$\therefore f(x)$ is continuous at all points except $x=0$.

If $f(x)$ is not continuous at $x=a$, we will say that $f(x)$ is discontinuous at $x=a$.

If we can draw the graph of the function $y=f(x)$, then the continuity of $f(x)$ at $x=a$ means that the graph is not "broken" near the point $(a, f(a))$ in the graph.

For $f(x)= \begin{cases}1, & x \geqslant 0 \\ 0, & x<0\end{cases}$

$f(x)$ has "broken" graph near $(0,1)$.

(2) $\quad f(x)= \begin{cases}1, & x \neq 0 \\ 0, & x=0\end{cases}$

Here $\lim _{x \rightarrow 0} f(x)$ exists & is equal to 1 .

But, $f(0)=0 \neq \lim _{x \rightarrow 0} f(x)$

Hence, $f(x)$ is discontinuous at $x=0$

Let's consider

$\quad f(x)= \begin{cases}x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{cases}$

Is $f$ continuous at $x=0$ ? It is difficult to draw the graph.

But the $\lim _{x \rightarrow 0} f(x)$ can be calculated as follows.

$|f(x)|=\left|x \sin \left(\frac{1}{x}\right)\right| \leqslant|x|\left|\sin \left(\frac{1}{x}\right)\right|$

$\therefore \quad\left|\sin \left(\frac{1}{x}\right)\right| \leqslant 1$

$\therefore 0 \leqslant\left|x \sin \left(\frac{1}{x}\right)\right| \leqslant|x|$

Since $\lim _{x \rightarrow 0} x=0=\lim _{x \rightarrow 0} 0$, by the Sandwich theorem, $\quad \lim _{x \rightarrow 0}\left|x \sin \left(\frac{1}{x}\right)\right|=0$

$\Rightarrow \quad \lim _{x \rightarrow 0} | x \sin \left(\frac{1}{x}\right)|=0$

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

So, $f(x)$ is continuous at $x=0$.

Let $f(x)$ be a real-valued function.

we say that $f(x)$ is differentiable at $x=a$ if the limit $\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ exists.

Note that for $f(x)$ to be differentiable at $x=a$ the function $f(x)$ must be defined in an open interval containing $x=a$.

As $h \rightarrow 0$ the point $Q \rightarrow P$, and the secant (ie. the line segment joining $P$ and Q) approaches the tangent line at the point $(a, f(a))$

So, $\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ is the slope of the tangent, provided the limit exists.

(1) $\quad f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=c$, a constant. $\forall x \in \mathbb{R}$.

For any $a \in \mathbb{R}, \quad f(a+h)-f(a)=c-c=0 \quad\forall h$

$\therefore \lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=0$

$\therefore f$ is differentiable at every a and $f^{\prime}(a)=0$.

(2) $f(x)=x$

$f^{\prime}(a) =\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$

$=\lim _{h \rightarrow 0} \frac{(a+h)-\alpha}{h}=\lim _{h \rightarrow 0} \frac{k^{\prime}}{k}=1$

So, for $f(x)=x, f^{\prime}(x)=1$ for all $x \in \mathbb{R}$

(3) $f(x)={x^2}$

$\frac{f(x+h)-f(x)}{h} =\frac{(x+h)^2-x^2}{h}, h \neq 0$

$=\frac{x^2+2 h x+h^2-x^2}{h}$

$=\frac{\not h(2 x+h)}{\not h}=2 x+h$

$\rightarrow 2 x \text { as } h \rightarrow 0$

$\therefore f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=2 x \text {. }$

Notation: $f^{\prime}(x)$ is also denoted by $\frac{d}{d x} f(x)$ is $\frac{d y}{d x}$, where $y=f(x)$.

So, $\frac{d}{d x}(c)=0$

$\frac{d}{d x}(x)=1$

$\frac{d}{d x}\left(x^2\right)=2 x$

Property: (1) Let $f(x)$ and $g(x)$ be differentiable at $x=a$. Then $f(x)+g(x)$ is also diff at $x=a$, and

$\frac{d}{d x}|_{x=a}(f(x)+g(x))$=

$\frac{d}{d x}|_{x=a} f(x)$

+$\frac{d}{d x}|_{x=a}g(x)=f^{\prime}(a)+g^{\prime}(a)$

Proof: Let $u(x)=f(x)+g(x)$

for $h \neq 0, \frac{u(a+h)-u(a)}{h}=[f(a+h)+g(a+h)]-[f(a)+g(a)]$

$\therefore \lim _{h \rightarrow 0} \frac{u(a+h)-u(a)}{h}= \lim _{h \rightarrow 0} \frac{f(a+h)}{h}-f(a)$

$+\lim _{h \rightarrow 0} \frac{g(a+h)-g(a)}{h}$

(by the sum Rule of limits)

ie. $u^{\prime}(a)=f^{\prime}(a)+g^{\prime}(a)$

(2) If $u(x)=c f(x)$ for some $c \in \mathbb{R}$, & $f^{\prime}(a)$ exists, then $u^{\prime}(a)$ exists and $u^{\prime}(a)=c f^{\prime}(a)$.

Proof: For $h \neq 0, \frac{u(a+h)-u(a)}{h}=\frac{c f(a+h)-c f(a)}{h}$ $=c\left[\frac{f(a+h)-f(a)}{h}\right]$

$\rightarrow c f^{\prime}(a)$ as $h \rightarrow 0$.

$\therefore u^{\prime}(a)=c f^{\prime}(a) \text {. }$

(3) If $u(x)=c_1 f(x)+c_2 g(x)$ and $f^{\prime}(a), g^{\prime}(a)$ exists then

$u^{\prime}(a)=c_1 f^{\prime}(a)+c_2 g^{\prime}(a)$.

Proof: $u^{\prime}(a)=\frac{d}{d x}|_{x=a}(c_1 f(x))$

$+\frac{d}{d x}|_{x=a}(c_2 g(x))$

$=c_1 f^{\prime}(a)+c_2 g^{\prime}(a)$

(by the previous the results).

Let $f(x)=x^2-3 x+2$

Calculate $f^{\prime}(3)$, if it exists.

Since $\frac{d}{d x}\left(x^2\right)=2 x, \frac{d}{d x}(x)=1, \frac{d}{d x}(2)=0$,

$f^{\prime}(x)=\frac{d}{d x}\left(x^2-3 x+2\right) =2 x-3(1)+0$

$=2x-3$

$\therefore f^{\prime}(3)=2(3)-3=3$ .