Continuity and Differentiability Exercise 01
Question:
Examine the continuity of f, where f is defined by $fx=\{\begin{array}{lllll}\mathrm{sin}x\mathrm{cos}x& & \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x=0& & \\ 1& & \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x=0& & \end{array}$
Answer:
 Find the limit of f(x) as x approaches 0.
lim f(x) = lim (sin x  cos x) as x approaches 0
 Evaluate the limit.
lim f(x) = lim (sin 0  cos 0) as x approaches 0
= lim (0  (1)) as x approaches 0
= 1
 Compare the limit with the value of f(0).
Since the limit of f(x) as x approaches 0 is 1 and the value of f(0) is 1, f is not continuous at x = 0.
Question:
Find the relationship between a and b, so that the function f(x) defined by $\{\begin{array}{lllll}ax+2& \text{,}& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 4& & \\ bx+4& \text{,}& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>4& & \end{array}$ is continuous at x=3
Answer:

First, we must determine the values of a and b. To do this, we need to determine the value of f(x) at x=3 when a and b are unknown.

For x ≤ 4, f(3) = a*3 + 2.

For x > 4, f(3) = b*3 + 4.

We can set these two equations equal to each other to solve for a and b: a3 + 2 = b3 + 4.

Subtracting 4 from both sides of the equation yields: a3  b3 = 2.

Dividing both sides of the equation by 3 yields: a  b = 2/3.

Therefore, the relationship between a and b is a = b  2/3.
Question:
Examine the following function for continuity. f(x)= x^2−25/x+5, x = 5
Answer:

Substitute x = 5 into the function: f(5) = 5^2  25/5 + 5 = 25  5 + 5 = 25

Check that the limit of the function as x approaches 5 is equal to f(5): lim x→5 f(x) = lim x→5 x^2  25/x + 5 = lim x→5 (x+5)(x5)/(x+5) = lim x→5 (x5) = 5  5 = 0

Since f(5) = 25 and the limit of the function as x approaches 5 is 0, the function is not continuous at x = 5.
Question:
Find the values of a and b such that the function defined by $\{\begin{array}{lllll}5& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 2& & & \\ ax+b& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}2<x<10& & & \\ 21& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\ge 10& & & \end{array}$ is a continuous function.
Answer:

Since the function must be continuous, the values of a and b must be such that the function is continuous at x = 2 and x = 10.

To make the function continuous at x = 2, the value of the function at x = 2 must be equal to the value of the function at x = 10.

Therefore, 5 = a2 + b and 21 = a10 + b.

Solving these equations, a = 8 and b = 13.
Question:
Discuss the continuity of the function defined by f(x)=∣x−5∣.
Answer:
Step 1: Define the function f(x) = x5
Step 2: Analyze the function to determine its continuity at x = 5.
Step 3: Calculate the lefthand limit of f(x) as x approaches 5.
Step 4: Calculate the righthand limit of f(x) as x approaches 5.
Step 5: Compare the lefthand limit and righthand limit.
Step 6: If the lefthand and righthand limits are equal, the function is continuous at x = 5.
Question:
Discuss the continuity of the function f, where f is defined by $\{\begin{array}{lllll}3& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}0\le x\le 1& & & \\ 4& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}1<x<3& & & \\ 5& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}3\le x\le 10& & & \end{array}$
Answer:

Define the domain of the function f: The domain of the function f is the set of all real numbers x such that 0 ≤ x ≤ 10.

Determine the continuity of the function f: The function f is continuous over its domain because the function is defined for all values in the domain and is continuous at each point in the domain.
Question:
$\{\begin{array}{lllll}x& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 1& & & \\ 5& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>1& & & \end{array}$ continuous at x=0? At x=1? At x=2?
Answer:
At x=0: The function is not defined, so it is not continuous.
At x=1: The function is continuous since the lefthand side and righthand side of the equation have the same value (5).
At x=2: The function is continuous since the lefthand side and righthand side of the equation have the same value (5).
Question:
Is the function f defined by $\{\begin{array}{lllll}{x}^{2}& \mathrm{sin}\left(\frac{1}{x}\right)& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x=0& & \\ 0& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x=0& & & \end{array}$ is a continuous function ?
Answer:
 Determine the domain of the function f.
Answer: The domain of the function f is x ≠ 0.
 Determine the limit of the function f as x approaches 0.
Answer: The limit of the function f as x approaches 0 is 0.
 Determine if the function f is continuous at x = 0.
Answer: Yes, the function f is continuous at x = 0 since the limit of the function f as x approaches 0 is equal to the value of the function at x = 0.
Question:
Examine the following functions for continuity.(i) f(x)=x−5 (ii) f(x)= 1/x−5,x=5 (iii) f(x)=x^2−25/x+5,x=−5 (iv) f(x)=∣x−5∣
Answer:
(i) f(x)=x−5 Continuous at all x
(ii) f(x)= 1/x−5,x=5 Continuous at x=5
(iii) f(x)=x^2−25/x+5,x=−5 Not continuous at x=5
(iv) f(x)=∣x−5∣ Continuous at all x
Question:
Find the values of k so that the function f is continuous at the indicated point: $\{\begin{array}{lllll}k{x}^{2}& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 2& & & \\ 3& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>2& & & \end{array}$
Answer:

Substitute the point x=2 into the equation to find the value of k when x is less than or equal to 2: k2 = kx2 = k(2)2 = 4

Substitute the point x=2 into the equation to find the value of k when x is greater than 2: k = 3
Therefore, the values of k so that the function f is continuous at the indicated point are k = 4 when x ≤ 2 and k = 3 when x > 2.
Question:
For what value of λ is the function defined by $\{\begin{array}{lllll}\lambda ({x}^{2}2x)& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 0& & & \\ 5& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>0& & & \end{array}$ continuous at x=0? What about continuity at x=1 ?
Answer:
For λ to be continuous at x=0, λ must equal 2x. Therefore, λ = 2x = 0 when x = 0.
For λ to be continuous at x=1, λ must equal 5. Therefore, λ = 5 when x = 1.
Question:
Find the values of k so that the function f is continuous at the indicated point: $\{\begin{array}{lllll}k{x}^{2}& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le \pi & & & \\ \mathrm{cos}x& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>\pi & & & \end{array}$ at x= π
Answer:

Determine the function f(x) at x=π: f(x) = kx2 if x ≤ π f(x) = cos(x) if x > π

Set the two functions equal to each other at x=π: kπ2 = cos(π)

Solve for k: k = cos(π) / π2
Question:
Find all points of discontinuity of f, where f is defined by $\{\begin{array}{lllll}{x}^{10}1& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 1& & & \\ {x}^{2}& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>1& & & \end{array}$
Answer:

Determine the domain of f. The domain of f is x ≤ 1 or x > 1.

Check for points of discontinuity at the endpoints of the domain. At x = 1, f is undefined, so there is a point of discontinuity at x = 1.

Check for points of discontinuity within the domain. There are no points of discontinuity within the domain, so the only point of discontinuity of f is x = 1.
Question:
Find all points of discontinuity of f, where f is defined by $\{\begin{array}{lllll}\mid x\mid +3=x+3& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 3& & & \\ 2x& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}3<x<3& & & \\ 6x& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\ge 3& & & \end{array}$
Answer:

Identify the points of discontinuity: The points of discontinuity of f are x=3 and x=3.

Determine the type of discontinuity at each point: At x=3, f has a jump discontinuity as the value of f jumps from 6x to 2x. At x=3, f has a removable discontinuity as the value of f is undefined.
Question:
Find all the points of discontinuity of f defined by f(x)=∣x∣−∣x+1∣.
Answer:
 Set f(x)=0
 Solve for x
 x=1
 Therefore, the point of discontinuity is x=1
Question:
Discuss the continuity of the function f, where f is defined by $\{\begin{array}{lllll}2x& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x<0& & & \\ 0& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}0\le x\le 1& & & \\ 4x& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>1& & & \end{array}$
Answer:

First, examine the domain of the function. The domain of f is all real numbers.

Next, examine the behavior of the function at the endpoints of the domain. For x < 0, the function f is equal to 2x. For x = 0, the function f is equal to 0. For 0 < x ≤ 1, the function f is equal to 0. For x > 1, the function f is equal to 4x.

Finally, examine the behavior of the function in between the endpoints. As x approaches 0 from the left, the function f approaches 0. As x approaches 0 from the right, the function f is equal to 0. As x approaches 1 from the left, the function f is equal to 0. As x approaches 1 from the right, the function f approaches 4.

Therefore, the function f is continuous on the domain.
Question:
Is the function defined by $\{\begin{array}{lllll}x+5& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 1& & & \\ x5& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>1& & & \end{array}$ a continuous function ?
Answer:
Answer: No, it is not a continuous function. The function is discontinuous at x = 1.
Question:
Prove that the function f(x)=x^n is continuous at x=n, where n is a positive integer.
Answer:
Proof:

First, we need to show that the limit of f(x) as x approaches n is equal to f(n).

We can use the definition of a limit to show that this is true.

Let ε > 0 be given.

We need to find a δ > 0 such that if x−n < δ, then f(x)−f(n) < ε.

Since f(x)=x^n, we can rewrite this as x^n−n^n < ε.

We can use the triangle inequality to simplify this expression to x−n·x^(n−1)+x^(n−2)n+…+xn+n^(n−1) < ε.

Since n is a positive integer, all of the terms in the absolute value are positive, so we can further simplify this expression to x−n·(n^(n−1)) < ε.

Now, we can set δ = ε/(n^(n−1)) to make sure that if x−n < δ, then f(x)−f(n) < ε.

Therefore, the limit of f(x) as x approaches n is equal to f(n).

Since the limit exists and is equal to f(n), f(x) is continuous at x=n.
Q.E.D.
Question:
Find the values of k so that the function f is continuous at the indicated point: $f\left(x\right)=\{\begin{array}{lllll}kx+1& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 5& & & \\ 3x5& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>5& & & \end{array}$ at x=5
Answer:
 Determine the function values at x=5 for both equations:
f(5) = k(5) +1 = 6k
f(5) = 3(5)  5 = 10  5 = 5
 Set the two function values equal to each other and solve for k:
6k = 5
k = 5/6
Question:
Find all points of discontinuity of f, where f is defined by $f\left(x\right)=\{\begin{array}{lllll}\frac{\mid x\mid}{x}& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x=0& & & \\ 0& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x=0& & & \end{array}$
Answer:
Answer: The point of discontinuity of f is x = 0.
Question:
Find all points of discontinuity of f, where $f\left(x\right)=\{\begin{array}{lllll}\frac{\mathrm{sin}x}{x}& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x<0& & & \\ x+1& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\ge 0& & & \end{array}$
Answer:
 The point of discontinuity of f(x) is x=0.
Question:
Discuss the continuity of the functionf, where f is defined by $f\left(x\right)=\{\begin{array}{lllll}2& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 1& & & \\ 2x& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}1<x\le 1& & & \\ 2& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>1& & & \end{array}$
Answer:

Identify the domain of the function. The domain of the function is all real numbers.

Determine the value of the function at the endpoints of the domain. At x = 1, f(x) = 2 and at x = 1, f(x) = 2.

Check if the function is continuous at the endpoints. Since the function is equal to the same value at both endpoints, it is continuous at the endpoints.

Check if the function is continuous in the interior of the domain. Since the function is a piecewise function, it is continuous in the interior of the domain.
Question:
Is the function defined by f(x)=x^2−sinx+5 continuous at x=π? Enter ‘0’ if yes else enter ‘1.’
Answer:
Answer: 0
Question:
Find the values of k so that the function f is continuous at the indicated point: $f\left(x\right)=\{\begin{array}{lllll}\frac{k\mathrm{cos}x}{\pi 2x}& x\ne \frac{\pi}{2}& & & \\ 3& x=\frac{\pi}{2}& & & \end{array}$ at x= 2π
Answer:

Substitute x= 2π into the equation: f(2π) = {kcos(2π) 2(2π) x ≠ 2π/2 3x= 2π/2}

Simplify the equation: f(2π) = {k  4π x ≠ 2π/2 3x= 2π/2}

Set the equation equal to 0: 0 = k  4π x ≠ 2π/2 3x= 2π/2

Solve for k: k = 4π
Question:
Examine the continuity of the function f(x)=2x^2−1 at x=3.
Answer:
Step 1: Calculate the limit of f(x) as x approaches 3:
lim x→3 f(x) = lim x→3 (2x2 − 1)
Step 2: Substitute x = 3 into the function f(x):
f(3) = 2(3)2 − 1
Step 3: Calculate the limit and the value of f(3):
lim x→3 f(x) = lim x→3 (2x2 − 1) = 17 f(3) = 2(3)2 − 1 = 17
Step 4: Compare the limit and the value of f(3):
Since the limit and the value of f(3) are equal, the function f(x) is continuous at x = 3.
Question:
Prove that the function f(x)=5x−3 is continuous at x=0, at x=−3 and at x=5.
Answer:
At x=0: f(0) = 5(0)  3 = 3
The lefthand limit at x=0 is 3 and the righthand limit at x=0 is 3, which means the function is continuous at x=0.
At x=3: f(3) = 5(3)  3 = 18
The lefthand limit at x=3 is 18 and the righthand limit at x=3 is 18, which means the function is continuous at x=3.
At x=5: f(5) = 5(5)  3 = 22
The lefthand limit at x=5 is 22 and the righthand limit at x=5 is 22, which means the function is continuous at x=5.
Question:
Find all points of discontinuity of f, where f is defined by $f\left(x\right)=\{\begin{array}{lllll}2x+3& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}\times \le 2& & & \\ 2x3& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>2& & & \end{array}$
Answer:
Answer: The point of discontinuity of f is x = 2.
Question:
Find all points of discontinuity of f, where f is defined by $f\left(x\right)=\{\begin{array}{lllll}x+1& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\ge 1& & & \\ {x}^{2}+1& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x<1& & & \end{array}$
Answer:
Step 1: Determine the domain of the function.
The domain of the function is all real numbers.
Step 2: Determine the points of discontinuity.
The point of discontinuity is x = 1.
Question:
Examine that sin x is a continuous function.
Answer:

First, define what is meant by a continuous function. A continuous function is one where the output value changes smoothly with the input value, without any sudden jumps or breaks.

Next, examine the function sin x. This function takes the absolute value of x, meaning that the output is always positive regardless of the sign of x.

Since this function is always increasing, it is continuous and therefore satisfies the definition of a continuous function.
Question:
Find all points of discontinuity of f, where f is defined by $f\left(x\right)=\{\begin{array}{lllll}\frac{x}{\leftx\right}& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x<0& & & \\ 1& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\ge 0& & & \end{array}$
Answer:

First, determine if the function has any points of discontinuity. Points of discontinuity are points at which the function is not continuous. This can be done by examining the function and seeing if there are any jumps or gaps in the graph of the function.

Next, examine the definition of the function to find any points of discontinuity. In this case, the function is defined by a piecewise function, which means that the function is defined differently for different values of x. The points of discontinuity are where the two pieces of the function join together. In this case, the point of discontinuity is x=0, since the two pieces of the function join together at this point.

Finally, check to make sure that the point of discontinuity is actually a point of discontinuity. To do this, take the limit of the function as x approaches 0 from the left and from the right. If the two limits are not equal, then the point of discontinuity is confirmed. In this case, the left limit is 0, and the right limit is 1, which means that the point of discontinuity is confirmed.
Therefore, the point of discontinuity of the given function is x=0.
Question:
Show that the function defined by g(x)=x−[x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.
Answer:
Step 1: First, let us define the function g(x). It is a function that takes a real number x as input and returns the difference between x and the greatest integer less than or equal to x.
Step 2: To show that the function is discontinuous at all integral points, let’s assume that x is an integral point. Then, the greatest integer less than or equal to x is x itself.
Step 3: Therefore, the function g(x) becomes g(x) = x  x = 0.
Step 4: Now, let’s consider the limit of the function g(x) as x approaches the integral point from the left. Since the greatest integer less than or equal to x is x itself, the limit of the function g(x) as x approaches the integral point from the left is 0.
Step 5: Now, let’s consider the limit of the function g(x) as x approaches the integral point from the right. Since the greatest integer less than or equal to x is x itself, the limit of the function g(x) as x approaches the integral point from the right is also 0.
Step 6: Since the limit of the function g(x) as x approaches the integral point from the left is not equal to the limit of the function g(x) as x approaches the integral point from the right, the function g(x) is discontinuous at all integral points.
Question:
Find all points of discontinuity of f, where f is defined by $f\left(x\right)=\{\begin{array}{lllll}{x}^{3}3& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x\le 2& & & \\ {x}^{2}+1& \phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}x>2& & & \end{array}$
Answer:

First, identify any places where the two parts of the definition of f are not equal. This is the point of discontinuity. In this case, the two parts of the definition of f are equal when x = 2.

Next, determine whether the point of discontinuity is removable or not. To do this, we need to evaluate the limit of f(x) as x approaches 2 from both the left and the right. If the two limits are equal, then the discontinuity is removable. Otherwise, it is not.
In this case, the lefthand limit is 3 and the righthand limit is 1. Therefore, the discontinuity is not removable, and the point of discontinuity is x = 2.