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)

For example, choice (a) should be False.

For example, choice (b) should be True.

For example, choice (c) should be False.

For example, choice (d) should be True.

For example, choice (e) should be False.

*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!*

*False*This function is not continuous at $x=2$.*True*This function is continuous at every point in its domain.*False*This function is not continuous at $x=2$.*True*This function is continuous at every point in its domain.*False*This function is not continuous at $x=2$.

At what value or values of $x$
is the function

For example, choice (a) should be True.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be False.

For example, choice (e) should be False.

$$f\left(x\right)=\left\{\begin{array}{cc}x+4\hfill & \text{if}x\le -1\text{}\hfill \\ {x}^{2}\hfill & \text{if}-1x1\text{}\hfill \\ 2-x\hfill & \text{if}x\ge 1\text{}\hfill \end{array}\right.$$

discontinuous?
(Zero or more options can be correct)
*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!*

*True**False**False**False**False*

At what value or values of $x$
is the function

$$f\left(x\right)=\left\{\begin{array}{cc}|x+1|-1\hfill & \text{if}x0\text{}\hfill \\ {x}^{2}+x\hfill & \text{if}0\le x1\text{}\hfill \\ 3-x\hfill & \text{if}1\le x\text{}\hfill \end{array}\right.$$

discontinuous? Exactly one option must be correct)
*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

$$f\left(x\right)=\left\{\begin{array}{cc}{x}^{3}\hfill & \text{if}x\le a\text{}\hfill \\ {x}^{2}\hfill & \text{if}xa\text{}\hfill \end{array}\right.$$

be continuous for all $x$? Exactly
one option must be correct)
*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)

For example, choice (a) should be True.

For example, choice (b) should be True.

For example, choice (c) should be False.

For example, choice (d) should be True.

*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!*

*True**True*The intermediate value theorem applies to continuous functions. As every differentiable function is continuous this argument is correct.*False*The fact that a function has a local maximum or minimum is independent of the question of the existence of roots.*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)

For example, choice (a) should be False.

For example, choice (b) should be True.

For example, choice (c) should be False.

For example, choice (d) should be True.

For example, choice (e) should be True.

*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!*

*False*L’Hôpital’s rule cannot be used here because, although $\underset{x\to \pi}{lim}x-\pi =0,$ it is not true that $\underset{x\to \pi}{lim}cosx=0$.*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 $\underset{x\to \infty}{lim}x=\infty =\underset{x\to \infty}{lim}{e}^{x}$.*False*L’Hôpital’s Rule can not be used here because although the denominator has limit 0, the numerator does not. That is, $\underset{x\to 1}{lim}ln\left(2x\right)=ln\left(2\right)\ne 0$.*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$$\underset{x\to 1}{lim}\frac{lnx}{x-1}=\underset{x\to 1}{lim}\frac{\frac{1}{x}}{1}=1.$$*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

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

*Incorrect.*

*Please try again.*

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,
$$\begin{array}{llll}\hfill \underset{x\to 0}{lim}\frac{6{e}^{x}-6x-6}{{x}^{2}}& =\underset{x\to 0}{lim}\frac{6{e}^{x}-6}{2x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\underset{x\to 0}{lim}\frac{3{e}^{x}-3}{x}\phantom{\rule{2em}{0ex}}& \hfill & \\ \multicolumn{4}{c}{\text{ApplyingL\u2019H\xf4pital\u2019sruleasecondtimewehave}}\\ \phantom{\rule{2em}{0ex}}\\ \hfill \underset{x\to 0}{lim}\frac{6{e}^{x}-6x-6}{{x}^{2}}& =\underset{x\to 0}{lim}\frac{3{e}^{x}}{1}=3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

*Incorrect.*

*Please try again.*

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,

$\underset{x\to 0}{lim}\frac{cosx-1}{x}=\underset{x\to 0}{lim}\frac{-sinx}{1}=0$

$\underset{x\to 0}{lim}\frac{cosx-1}{x}=\underset{x\to 0}{lim}\frac{-sinx}{1}=0$

*Incorrect.*

*Please try again.*

Try using L’Hôpital’s Rule.

What is the value of the limit $\underset{x\to \infty}{lim}\left(\right.1+\frac{1}{x}{\left)\right.}^{x}$
?
Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is correct!*

*Choice (c) is incorrect*

*Choice (d) is incorrect*