Continuity and Discontinuity Functions Worksheet With Answers

Exercise 1

Study the following functions and determine if they are continuous. If not, state where the discontinuities exist and what type they are:

1 $f(x) = \frac{5}{x^4 - 16}$

2 $f(x) = \frac{x - 7}{x^3 - x^2 - 11x + 3}$

3 $f(x) = \left\{\begin{matrix} x^2 - 1 if x \leq 0 \\ 2x - 3, if x>0 \end{matrix}\right$

4 $f(x) = \left\{\begin{matrix} x^2 - 1 if x \leq 0 \\2x - 3, if x>0 \end{matrix}\right$

5 $f(x) = \left\{\begin{matrix} \frac{1}{x} if x < 1 \\\sqrt{x + 1}, if x\geq 1 \end{matrix}\right$

6 $f(x) = \left\{\begin{matrix} \frac{e^x}{e^x + 1} if x \leq 0 \\x^2 + 1, if x> 1 \end{matrix}\right$

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Exercise 2

Determine if the following function is continuous at x = 0.

$f(x) = \left\{\begin{matrix} x ^ {\frac{1}{x}}, if x > 0 \\0, if x = 0 \end{matrix}\right$

Exercise 3

Determine if the following function is continuous on (0,3). If not, state where the discontinuities exist and what type they are:

$f(x) = \left\{\begin{matrix} x^2, if0 < x < 1 \\0, if 1 \leq x < 2, x - 1, if 2 \leq x < 3 \end{matrix}\right$

Exercise 4

Are the following functions continuous at x = 0?

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

Exercise 5

Given the function:

$f(x) = \left\{\begin{matrix} \frac{x^2 - 25}{x - 5}, if x \neq 5 \\0, if x= 5\end{matrix}\right$

1 Prove that f(x) is not continuous at x = 5.

2Is there a continuous function which coincides with f(x) for all values with the exception x = 5? If so, determine the function.

Exercise 6

Determine if the following function is continuous. If not, state where the discontinuities exist or why the function is not continuous:
$f(x) = \frac{x + 1}{|x|}$

Exercise 7

Determine if the following function is continuous at x = 0.

$f(x) = \left\{\begin{matrix} x \cdot sin \frac{1}{x} if x \neq 0 \\0, if x = 0 \end{matrix}\right$

Exercise 8

Determine the value of a to make the following function continuous.

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

Exercise 9

The function defined by:

$f(x) = \left\{\begin{matrix} \sqrt{ax} if 0 \leq x \leq 8 \\\frac{x^2 - 32}{x - 4}, if x> 8 \end{matrix}\right$

is continuous on [0, ∞).

Determine the value of a that would make this statement true.

Solution of exercise 1

Study the following functions and determine if they are continuous. If not, state where the discontinuities exist:

1 $f(x) = \frac{5}{x^4 - 16}$

The function is continuous at all points of its domain.

$D = R - {-2, 2}$

D = R − {−2,2}

The function has two points of discontinuity at x = −2 and x = 2.

2 $f(x) = \frac{x - 7}{x^3 - x^2 - 11x + 3}$

The function is continuous at R with the exception of the values that annul the denominator. If this is equal to zero and the equation is solved, the discontinuity points will be obtained.

$x^3 - x^2 - 11x + 3 = 0$

x = −3; and by solving the quadratic equation: $x = 2 - \sqrt{3}$ and $x = 2 + \sqrt{3}$ are also obtained

The function has three points of discontinuity at $x = 3$ , $x = 2 - \sqrt{3}$ and $x = 2 + \sqrt{3}$ .

3 $f(x) = \left\{\begin{matrix} x^2 - 1 if x \leq 0 \\ 2x - 3, if x>0 \end{matrix}\right$

$f(2) = 3$

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

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

The function is continuous.

4 $f(x) = \left\{\begin{matrix} x^2 - 1 if x \leq 0 \\2x - 3, if x>0 \end{matrix}\right$

$f(0) = -1$

$\lim _ {x \rightarrow 0^{-}} (x^2 - 1) = -1$

$\lim _ {x \rightarrow 0^{+}} (2x - 3) = -3$

The function has a jump discontinuity at x = 0 .

5 $f(x) = \left\{\begin{matrix} \frac{1}{x} if x < 1 \\\sqrt{x + 1}, if x\geq 1 \end{matrix}\right$

$f(1) = \sqrt{2}$

$\lim_{x \rightarrow 1^{-}} (\frac{1}{x}) = 1$

$\lim_{x \rightarrow 1^{+}} \sqrt{x + 1} = \sqrt{2}$

The function has a jump discontinuity at x = 1 .

6 $f(x) = \left\{\begin{matrix} \frac{e^x}{e^x + 1} if x \leq 0 \\x^2 + 1, if x> 1 \end{matrix}\right$

$f(0) = \frac{1}{2}$

$\lim_{x \rightarrow 0^{-}} (\frac {e^x}{e^x + 1}) = \frac{1}{2}$

$\lim_{x \rightarrow 0^{+}} x^2 + 1 = 1$

The function has a jump discontinuity at x = 1/2 .

Solution of exercise 2

Determine if the following function is continuous at x = 0.

$f(x) = \left\{\begin{matrix} x ^ {\frac{1}{x}}, if x > 0 \\0, if x = 0 \end{matrix}\right$

$f(0) = 0$

$\lim_{x \rightarrow 0^{+}} x^ {\frac{1}{x}} = 0 ^ {\frac{1}{0^{+}}} = 0 ^{\infty} = 0$

$\lim_{x \rightarrow 0^{-}} x^ {\frac{1}{x}}$

At x = 0, there is an essential discontinuity.

Solution of exercise 3

Determine if the following function is continuous on (0,3). If not, state where the discontinuities exist and what type they are:

$f(x) = \left\{\begin{matrix} x^2, if0 < x < 1 \\0, if 1 \leq x < 2, x - 1, if 2 \leq x < 3 \end{matrix}\right$

$f(1) = 0$

$\lim_{x \rightarrow 1^{-}} x^2 = 1$

$\lim_{x \rightarrow 1^{+}} 0 = 0$

At x = 1, there is a jump discontinuity.

$f(2) = 1$

$\lim_{x \rightarrow 2^{-}} 0 = 0$

$\lim_{x \rightarrow 2^{+}} x - 1 = 1$

At x = 2, there is a jump discontinuity.

Solution of exercise 4

Are the following functions continuous at x = 0?

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

$f(0) = 2 ^{-0} = \frac{1}{2^{0}} = 1$

$\lim_{x \rightarrow 0^{-}} 2^{-x} = 2 ^{0^{-}} = 2^0 = 1$

$\lim_{x \rightarrow 0^{+}} 2^{-x} = 2^{-0^{+}} = \frac{1}{2^0} = 1$

The function is continuous at x = 0.

Solution of exercise 5

Given the function:

$f(x) = \left\{\begin{matrix} \frac{x^2 - 25}{x - 5}, if x \neq 5 \\0, if x= 5\end{matrix}\right$

1 Prove that f(x) is not continuous at x = 5.

$f(5) = 0$

$\lim_{x \rightarrow 5} \frac{x^2 - 25}{x - 5} = \frac{0}{0}$

Solve the indeterminate form.

$\lim_{x \rightarrow 5} \frac{(x + 5) (x - 5)}{x - 5} = \lim_{x \rightarrow 5} (x + 5) = 10$

f (x) is not continuous at x = 5 because:

$\lim_{x \rightarrow 5} f(x) \neq f(5)$

2 Is there a continuous function which coincides with f(x) for all values with the exception x = 5? If so, determine the function.

$\lim_{x \rightarrow 5} f(x) = f(5) = 10$

the function would be continuous, then the function is redefined:

$f(x) = \left\{\begin{matrix} \frac{x^2 - 25}{x - 5}, if x \neq 5 \\0, if x= 5\end{matrix}\right$

Solution of exercise 6

Determine if the following function is continuous. If not, state where the discontinuities exist or why the function is not continuous:

$f(x) = \frac{x + 1}{|x|}$

The function f(x) is continuous for x ≠ 0. Therefore, study the continuity at x = 0.

$\lim_{x \rightarrow 0 ^{-}} \frac{x + 1}{|x|} = \lim_{x \rightarrow 0^{-}} \frac{x + 1}{-x} = \lim_{x \rightarrow 0^{-}}(-1 - \frac{1}{x}) = -1 - \frac{1}{0^{-}} = \infty$

$\lim_{x \rightarrow 0 ^{+}} \frac{x + 1}{|x|} = \lim_{x \rightarrow 0^{+}} \frac{x + 1}{x} = \lim_{x \rightarrow 0^{+}}(-1 - \frac{1}{x}) = -1 - \frac{1}{0^{+}} = \infty$