Chapters

Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 5
Exercise 6
Exercise 7
Exercise 8
Exercise 9
Exercise 10
Exercise 11
Exercise 12
Solution of exercise 1
Solution of exercise 2
Solution of exercise 3
Solution of exercise 4
Solution of exercise 5
Solution of exercise 6
Solution of exercise 7
Solution of exercise 8
Solution of exercise 9
Solution of exercise 10
Solution of exercise 11
Solution of exercise 12

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Exercise 1

Prove that the function $\text{[math]}$ intersects the x-axis on the interval $\text{[math]}$ . Can the same be said for the function: $\text{[math]}$ ?

Exercise 2

Given the function:

$\text{[math]}$

Can it be said that f(x) is bounded in the interval $\text{[math]}$ ?

Exercise 3

Given the function $\text{[math]}$ . Can it be said that the function exists for all values in the interval $\text{[math]}$ ?

Exercise 4

Prove that the equation: $\text{[math]}$ , has at least one solution $\text{[math]}$ such that $\text{[math]}$ .

Exercise 5

Given the function $\text{[math]}$ . Can it be said that there is at least one point, c, inside the interval $\text{[math]}$ which verifies that $\text{[math]}$ ?

Exercise 6

Prove that the polynomial function $\text{[math]}$ has a value of zero between $\text{[math]}$ and $\text{[math]}$ .

Exercise 7

Prove that the equation $\text{[math]}$ has at least one real solution.

Exercise 8

Prove that there is a real number, x, such that $\text{[math]}$ .

Exercise 9

Given the function:

$\text{[math]}$

Prove that there is a point in the open interval $\text{[math]}$ in which the function $\text{[math]}$ has a value of $\text{[math]}$ .

Exercise 10

Given the function $\text{[math]}$ , determine if it is bounded superiorly and inferiorly in the interval $\text{[math]}$ and indicate if it reaches its maximum and minimum values within this interval.

Exercise 11

Prove that the function $\text{[math]}$ is continuous at $\text{[math]}$ and prove that there exists at least one real root of the equation $\text{[math]}$ .

Exercise 12

f and g are two continuous functions in $\text{[math]}$ and such that $\text{[math]}$ and $\text{[math]}$ . Prove the existance of c withinin $\text{[math]}$ such that $\text{[math]}$ .

Solution of exercise 1

Prove that the function $\text{[math]}$ intersects the x-axis on the interval $\text{[math]}$ . Can the same be said for the function: $\text{[math]}$ ?

The first function is continuous at $\text{[math]}$ .

$\text{[math]}$ .

Since it verifies the intermediate value theorem, at least one c belongs to the interval $\text{[math]}$ and intersects the x-axis.

We cannot confirm the same of the second function because it is not continuous at $\text{[math]}$ .

Solution of exercise 2

Given the function:

$\text{[math]}$

Can it be said that f(x) is bounded in the interval $\text{[math]}$ ?

Since f(x) is not continuous at $\text{[math]}$ , the function is not continuous in the closed interval $\text{[math]}$ , as a result, it cannot be said that the function is bounded in that interval.

Solution of exercise 3

Given the function $\text{[math]}$ . Can it be said that the function exists for all values in the interval $\text{[math]}$ ?

$\text{[math]}$ $\text{[math]}$

The function is continuous at $\text{[math]}$ since it is a polynomial function.

It is in the interval $\text{[math]}$ as it is verified that $\text{[math]}$ and $\text{[math]}$ .

Since it verifies the intermediate value theorem, the function exists at all values in the interval $\text{[math]}$ .

Solution of exercise 4

Prove that the equation: $\text{[math]}$ , has at least one solution $\text{[math]}$ such that $\text{[math]}$ .

f(x) is continuous in $\text{[math]}$

$\text{[math]}$

Since it verifies the Bolzano's Theorem, there is c $\text{[math]}$ such that:

$\text{[math]}$

Therefore there is at least one real solution to the equation $\text{[math]}$ .

Solution of exercise 5

Given the function $\text{[math]}$ . Can it be said that there is at least one point, c, inside the interval $\text{[math]}$ which verifies that $\text{[math]}$ ?

f(x) is continuous in $\text{[math]}$ .

$\text{[math]}$

The Bolzano theorem cannot be applied because it does not change sign.

Solution of exercise 6

Prove that the polynomial function $\text{[math]}$ has a value of zero between $\text{[math]}$ and $\text{[math]}$ .

$\text{[math]}$ is a polynomial and therefore is continuous in the interval $\text{[math]}$ .

$\text{[math]}$

$\text{[math]}$ .

There is a $\text{[math]}$ such that $\text{[math]}$

Solution of exercise 7

Prove that the equation $\text{[math]}$ has at least one real solution.

The function is continuous in the interval $\text{[math]}$ .

$\text{[math]}$ .

Since it verifies Bolzano's theorem, there is $\text{[math]}$ such that:

$\text{[math]}$

Therefore there is at least one real solution to the equation $\text{[math]}$ .

Solution of exercise 8

Prove that there is a real number, x, such that $\text{[math]}$ .

Consider the function $\text{[math]}$ .

It is continuous at $\text{[math]}$ .

$\text{[math]}$

There is a $\text{[math]}$ such that:

$\text{[math]}$

Therefore, there is at least one real solution to the equation $\text{[math]}$ .

Solution of exercise 9

Given the function:

$\text{[math]}$

Prove that there is a point in the open interval $\text{[math]}$ in which the function $\text{[math]}$ has a value of $\text{[math]}$ .

The exponential function is positive at $\text{[math]}$ , therefore the denominator of the function cannot be annulled.

There is only doubt of the continuity at $\text{[math]}$ , which is out of the interval being studied, therefore $\text{[math]}$ is continuous in $\text{[math]}$ .

Consider the function g defined by $\text{[math]}$ .

g is continuous on the interval $\text{[math]}$ .

$\text{[math]}$ $\text{[math]}$

Since it verifies the intermediate value theorem, there is a $\text{[math]}$ such that:

$\text{[math]}$

Solution of exercise 10

The function is continuous in the interval $\text{[math]}$ , as a result, it can be affirmed that it is bounded in that interval.

As well as being continuous in the interval $\text{[math]}$ , it has fulfilled the extreme value theorem, which affirms that it attains at least one maximum and absolute minimum in the interval $\text{[math]}$ .

Solution of exercise 11

Prove that the function $\text{[math]}$ is continuous at $\text{[math]}$ and prove that there exists at least one real root of the equation $\text{[math]}$ .

The function is continuous since it is the sum of continuous functions.

$\text{[math]}$

Since it verifies the intermediate value theorem, there is a $\text{[math]}$ such that:

$\text{[math]}$

Therefore, there is at least one real solution to the equation $\text{[math]}$ .

Solution of exercise 12

f and g are two continuous functions in $\text{[math]}$ and such that $\text{[math]}$ and $\text{[math]}$ . Prove the existance of c withinin $\text{[math]}$ such that $\text{[math]}$ .

h is the function defined by $\text{[math]}$ .

Since f and g are continuous in $\text{[math]}$ , the function h also is.

$\text{[math]}$ $\text{[math]}$

Since it verifies the intermediate value theorem, there is a $\text{[math]}$ such that:

$\text{[math]}$

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

Theory

Zero Over Zero

Calculating Limits

Continuity

Continuous Function

Continuity on a Closed Interval

Discontinuity

Essential Discontinuity

Intermediate Value Theorem

Jump Discontinuity

Limit

One-Sided Limit

Indeterminate Forms

Zero Times Infinity

Infinity Minus Infinity

Infinity over Infinity

Extreme Value Theorem

Bolzano’s Theorem

Limits at Infinity

Infinite Limit

Limit of a Logarithmic Function

Removable Discontinuity

Limit of an Exponential Function

Limit Rules

Properties of Infinity

One to the Power of Infinity

Limits and Dividing by Zero: Theory and Examples

Formulas

Continuity Formulas

Limit Formulas

Exercises

Limits Worksheet

Continuity Problems

Limit Problems

Intermediate Value Theorem Problems

Continuity Worksheet

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Don

November 2025

Derivative of denominator is wrong

Vanessa

November 2025

Good catch—thanks for pointing that out! We’ll double-check the derivative in that section and make any necessary corrections. Really appreciate you taking the time to flag it. 👍

Abbas Adamu

July 2025

Thank you

Vanessa

August 2025

Thank you Abbas! Good luck with your studies!

Mark Morton

July 2025

With regard to the Zero Over a Number item, is there a mis-statement? It’s immediately followed by “If a number is divided by zero which means that the numerator is zero and the denominator is the number, then the result is zero.”

Vanessa

July 2025

Hi Mark,

You’re absolutely right to raise the question — there does appear to be a misstatement in that sentence. The phrase “If a number is divided by zero, which means that the numerator is zero and the denominator is the number…” is indeed misleading and should be corrected.

To clarify:

Zero divided by a number (e.g. 0 ÷ 5) equals 0.

A number divided by zero (e.g. 5 ÷ 0) is undefined.

We’ll update the sentence to reflect the correct mathematical explanation. We appreciate you catching that and helping us improve the accuracy of the content!

Janice

July 2025

There is more than one size of infinity, though. What if you multiply the infinity of the whole numbers (Aleph-0) by the infinity of the real numbers (fraktur-c)?

Piyash Mia

May 2025

Thanks a lot to you for this essentiol article.

Vanessa

July 2025

Hi Piyash! Thanks for your comment, great to hear that you found this useful!

Jayaraman

May 2025

Very nice