## MATH1901 Quizzes

Quiz 6: Continuous functions
Question 1 Questions
Which of the following functions are continuous functions on the whole of their domain? (Tick each one that is continuous.) (Zero or more options can be correct)
 a) $f:ℝ\to ℝ;f\left(x\right)=\frac{2x+3}{x-2}$ if $x\ne 2$ and $f\left(x\right)=1$ if $x=2$. b) $f:\left(-\infty ,0\right)\to ℝ;f\left(x\right)=\frac{2x+3}{x-2}$ c) $f:\left(0,\infty \right)\to ℝ;f\left(x\right)=\frac{2x+3}{x-2}$ if $x\ne 2$ and $f\left(x\right)=1$ if $x=2$. d) $f:\left(-\infty ,2\right)\to ℝ;f\left(x\right)=\frac{2x+3}{x-2}$ e) $f:\left[2,\infty \right)\to ℝ;f\left(x\right)=\frac{2x+3}{x-2}$ if $x\ne 2$ and $f\left(x\right)=1$ if $x=2$.

There is at least one mistake.
For example, choice (a) should be False.
This function is not continuous at $x=2$.
There is at least one mistake.
For example, choice (b) should be True.
This function is continuous at every point in its domain.
There is at least one mistake.
For example, choice (c) should be False.
This function is not continuous at $x=2$.
There is at least one mistake.
For example, choice (d) should be True.
This function is continuous at every point in its domain.
There is at least one mistake.
For example, choice (e) should be False.
This function is not continuous at $x=2$.
Correct!
1. False This function is not continuous at $x=2$.
2. True This function is continuous at every point in its domain.
3. False This function is not continuous at $x=2$.
4. True This function is continuous at every point in its domain.
5. False This function is not continuous at $x=2$.
At what value or values of $x$ is the function
discontinuous? (Zero or more options can be correct)
 a) $-1$ b) 0 c) 1 d) 2 e) $f\left(x\right)$ is continuous everywhere

There is at least one mistake.
For example, choice (a) should be True.
There is at least one mistake.
For example, choice (b) should be False.
There is at least one mistake.
For example, choice (c) should be False.
There is at least one mistake.
For example, choice (d) should be False.
There is at least one mistake.
For example, choice (e) should be False.
Correct!
1. True
2. False
3. False
4. False
5. False
At what value or values of $x$ is the function
discontinuous? Exactly one option must be correct)
 a) $-1$, $0$ and $1$ b) 1 c) 0 d) $0$ and $1$ e) $f\left(x\right)$ is continuous everywhere

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is correct!
For what values of $a$ will the function
be continuous for all $x$? Exactly one option must be correct)
 a) $1$ and 2 b) $\sqrt{2}$ and $1$ c) $-1$ and $1$ d) $0$ and 1 e) $-1$ and $0$

Choice (a) is incorrect
As ${2}^{3}\ne {2}^{2}$ the function is not continuous if $a=2$.
Choice (b) is incorrect
As ${\left(\sqrt{2}\right)}^{3}\ne {\left(\sqrt{2}\right)}^{2}$ the function is not continuous if $a=\sqrt{2}$.
Choice (c) is incorrect
As ${\left(-1\right)}^{3}\ne {\left(-1\right)}^{2}$ the function is not continuous if $a=-1$.
Choice (d) is correct!
For the function to be continuous at $x=a$, we must have or ${a}^{3}={a}^{2}$. Hence $a=0$ or 1.
Choice (e) is incorrect
As ${\left(-1\right)}^{3}\ne {\left(-1\right)}^{2}$ the function is not continuous if $a=-1$.
Which of the following are correct proofs, using the Intermediate Value Theorem, that the polynomial $f\left(x\right)={x}^{3}+2{x}^{2}-1$ has at least one root? (More than one answer may be correct.) (Zero or more options can be correct)
 a) The function $f\left(x\right)$ is continuous for all real $x$ and $f\left(0\right)=-1$ and $f\left(1\right)=2$. Therefore, since $f\left(0\right)<0, by the Intermediate Value Theorem there is a number $c$ in $\left(0,1\right)$ such that $f\left(c\right)=0$. In other words, $c$ is a root of ${x}^{3}+2{x}^{2}-1=0$. b) The function $f\left(x\right)$ is differentiable for all real $x$ and $f\left(0\right)=-1$ and $f\left(2\right)=15$. Therefore, since $f\left(0\right)<0, by the Intermediate Value Theorem there is a number $c$ in $\left(0,2\right)$ such that $f\left(c\right)=0$. In other words, $c$ is a root of ${x}^{3}+2{x}^{2}-1=0$. c) As ${f}^{\prime }\left(x\right)=3{x}^{2}+4x$, the critical points of $f\left(x\right)$ occur at $x=0$ and $x=-\frac{4}{3}$. As $x=-\frac{4}{3}$ is a maximum $f\left(x\right)$ must have a root close to $x=-\frac{4}{3}$. d) The function $f\left(x\right)$ is continuous for all real $x$ and $f\left(0\right)=-1$ and $f\left(2\right)=15$. Therefore, since $f\left(0\right)<0, by the Intermediate Value Theorem there is a number $c$ in $\left(0,2\right)$ such that $f\left(c\right)=0$. In other words, $c$ is a root of ${x}^{3}+2{x}^{2}-1=0$.

There is at least one mistake.
For example, choice (a) should be True.
There is at least one mistake.
For example, choice (b) should be True.
The intermediate value theorem applies to continuous functions. As every differentiable function is continuous this argument is correct.
There is at least one mistake.
For example, choice (c) should be False.
The fact that a function has a local maximum or minimum is independent of the question of the existence of roots.
There is at least one mistake.
For example, choice (d) should be True.
Correct!
1. True
2. True The intermediate value theorem applies to continuous functions. As every differentiable function is continuous this argument is correct.
3. False The fact that a function has a local maximum or minimum is independent of the question of the existence of roots.
4. True
In which of the following cases can you correctly use L’Hôpital’s rule to evaluate the limit? (Zero or more options can be correct)
 a) $\underset{x\to \pi }{lim}\frac{cosx}{x-\pi }$ b) $\underset{x\to \infty }{lim}x{e}^{-x}$ c) $\underset{x\to 1}{lim}\frac{ln\left(2x\right)}{lnx}$ d) $\underset{x\to 1}{lim}\frac{lnx}{x-1}$ e) $\underset{x\to 0}{lim}\frac{tanx}{x}$

There is at least one mistake.
For example, choice (a) should be False.
L’Hôpital’s rule cannot be used here because, although $\underset{x\to \pi }{lim}x-\pi =0,$ it is not true that $\underset{x\to \pi }{lim}cosx=0$.
There is at least one mistake.
For example, choice (b) should be True.
L’Hôpital’s Rule can be used.
$\underset{x\to \infty }{lim}x{e}^{-x}=\underset{x\to \infty }{lim}\frac{x}{{e}^{x}}=\underset{x\to \infty }{lim}\frac{1}{{e}^{x}}=0.$
Note that $\underset{x\to \infty }{lim}x=\infty =\underset{x\to \infty }{lim}{e}^{x}$.
There is at least one mistake.
For example, choice (c) should be False.
L’Hôpital’s Rule can not be used here because although the denominator has limit 0, the numerator does not. That is, $\underset{x\to 1}{lim}ln\left(2x\right)=ln\left(2\right)\ne 0$.
There is at least one mistake.
For example, choice (d) should be True.
Since $\underset{x\to 1}{lim}lnx=0=\underset{x\to 1}{lim}x-1$ we can apply L’Hôpital’s Rule. This shows that
$\underset{x\to 1}{lim}\frac{lnx}{x-1}=\underset{x\to 1}{lim}\frac{\frac{1}{x}}{1}=1.$
There is at least one mistake.
For example, choice (e) should be True.
As $\underset{x\to 0}{lim}tanx=0=\underset{x\to 0}{lim}x$ we can apply L’Hôpital’s Rule, to find that
$\underset{x\to 0}{lim}\frac{tanx}{x}=\underset{x\to 0}{lim}\frac{{sec}^{2}x}{1}=\underset{x\to 0}{lim}\frac{1+{tan}^{2}x}{1}=0.$
Correct!
1. False L’Hôpital’s rule cannot be used here because, although $\underset{x\to \pi }{lim}x-\pi =0,$ it is not true that $\underset{x\to \pi }{lim}cosx=0$.
2. True L’Hôpital’s Rule can be used.
$\underset{x\to \infty }{lim}x{e}^{-x}=\underset{x\to \infty }{lim}\frac{x}{{e}^{x}}=\underset{x\to \infty }{lim}\frac{1}{{e}^{x}}=0.$
Note that $\underset{x\to \infty }{lim}x=\infty =\underset{x\to \infty }{lim}{e}^{x}$.
3. False L’Hôpital’s Rule can not be used here because although the denominator has limit 0, the numerator does not. That is, $\underset{x\to 1}{lim}ln\left(2x\right)=ln\left(2\right)\ne 0$.
4. True Since $\underset{x\to 1}{lim}lnx=0=\underset{x\to 1}{lim}x-1$ we can apply L’Hôpital’s Rule. This shows that
$\underset{x\to 1}{lim}\frac{lnx}{x-1}=\underset{x\to 1}{lim}\frac{\frac{1}{x}}{1}=1.$
5. True As $\underset{x\to 0}{lim}tanx=0=\underset{x\to 0}{lim}x$ we can apply L’Hôpital’s Rule, to find that
$\underset{x\to 0}{lim}\frac{tanx}{x}=\underset{x\to 0}{lim}\frac{{sec}^{2}x}{1}=\underset{x\to 0}{lim}\frac{1+{tan}^{2}x}{1}=0.$
Using L’Hôpital’s Rule find the value of $\underset{x\to 0}{lim}\frac{2\sqrt{1+3x}-2}{x}.$ Type your answer into the box.

Correct!
Since $\underset{x\to 0}{lim}2\sqrt{1+3x}-2=0=\underset{x\to 0}{lim}x$ we can apply L’Hôpital’s Rule. This gives
$\underset{x\to 0}{lim}\frac{2\sqrt{1+3x}-2}{x}=\underset{x\to 0}{lim}\frac{\frac{3}{\sqrt{1+3x}}}{1}=\underset{x\to 0}{lim}\frac{3}{\sqrt{1+3x}}=3.$

By L’Hôpital’s rule,

$\underset{x\to 0}{lim}\frac{\sqrt{1+3x}-1}{x}=\underset{x\to 0}{lim}\frac{\frac{3}{2\sqrt{1+3x}}}{1}$

Using L’Hôpital’s Rule find the value of $\underset{x\to 0}{lim}\frac{6{e}^{x}-6x-6}{{x}^{2}}.$ Type your answer into the box.

Correct!
Notice that we can apply L’Hôpital’s Rule because $\underset{x\to 0}{lim}6{e}^{x}-6x-6=0=\underset{x\to 0}{lim}{x}^{2}$. Therefore,

You need to use L’Hôpital’s Rule more than once.
Find the value of the limit $\underset{x\to 0}{lim}\frac{cosx-1}{x}.$ Type your answer into the box.

Correct!
By L’Hôpital’s Rule,
$\underset{x\to 0}{lim}\frac{cosx-1}{x}=\underset{x\to 0}{lim}\frac{-sinx}{1}=0$
What is the value of the limit $\underset{x\to \infty }{lim}\left(\right1+\frac{1}{x}{\left)\right}^{x}$ ? Exactly one option must be correct)
 a) $1$ b) $e$ c) $\infty$ d) The limit does not exist.