## MATH1001 Quizzes

Quiz 10: Limits, injectivity, inverse functions
Question 1 Questions
Check all of the following functions which are injective. (Zero or more options can be correct)
 a) $f\left(x\right)=lnx$ on domain $\left(0,\infty \right)$ b) $f\left(x\right)=coshx$ on domain $\left[0,\infty \right)$ c) $f\left(x\right)={x}^{3}+1$ on domain $ℝ$ d) $f\left(x\right)=sinx$ on domain $ℝ$

There is at least one mistake.
For example, choice (a) should be True.
The function is injective on this domain because its graph passes the horizontal line test, or alternatively, its derivative ${f}^{\prime }\left(x\right)=1∕x$ is positive, indicating an increasing (hence injective) function.
There is at least one mistake.
For example, choice (b) should be True.
The function is injective on this domain because its derivative ${f}^{\prime }\left(x\right)=sinhx$ is positive for all $x$ in $\left(0,\infty \right)$, indicating an increasing (hence injective) function. Note that the domain used here is not the natural domain, and has been chosen to make $cosh$ injective.
There is at least one mistake.
For example, choice (c) should be True.
The function is injective on this domain because its graph passes the horizontal line test.
There is at least one mistake.
For example, choice (d) should be False.
The function is not injective on this domain because, for example, the distinct input values $\pi$ and $2\pi$ give the same output value $0$.
Correct!
1. True The function is injective on this domain because its graph passes the horizontal line test, or alternatively, its derivative ${f}^{\prime }\left(x\right)=1∕x$ is positive, indicating an increasing (hence injective) function.
2. True The function is injective on this domain because its derivative ${f}^{\prime }\left(x\right)=sinhx$ is positive for all $x$ in $\left(0,\infty \right)$, indicating an increasing (hence injective) function. Note that the domain used here is not the natural domain, and has been chosen to make $cosh$ injective.
3. True The function is injective on this domain because its graph passes the horizontal line test.
4. False The function is not injective on this domain because, for example, the distinct input values $\pi$ and $2\pi$ give the same output value $0$.
If $f\left(x\right)={x}^{2}+1$ on domain $\left(0,\infty \right)$ and $g\left(x\right)={e}^{x}$ on domain $ℝ$, find the formula for the inverse function of the composite function $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ and find its range. Exactly one option must be correct)
 a) ${\left(f\circ g\right)}^{-1}\left(x\right)=\sqrt{ln\left(x-1\right)},\phantom{\rule{1em}{0ex}}$ range $ℝ$ b) ${\left(f\circ g\right)}^{-1}\left(x\right)=-\sqrt{ln\left(x\right)-1},\phantom{\rule{1em}{0ex}}$ range $\left(0,\infty \right)$ c) ${\left(f\circ g\right)}^{-1}\left(x\right)=\frac{1}{2}ln\left(x-1\right),\phantom{\rule{1em}{0ex}}$ range $ℝ$ d) ${\left(f\circ g\right)}^{-1}\left(x\right)={e}^{-2x+1},\phantom{\rule{1em}{0ex}}$ range $\left(0,\infty \right)$ e) ${\left(f\circ g\right)}^{-1}\left(x\right)=\frac{1}{2}ln\left(x-1\right),\phantom{\rule{1em}{0ex}}$ range $\left(1,\infty \right)$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The composite function has formula $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)={e}^{2x}+1.$ It has domain $ℝ$ and range $\left(1,\infty \right)$. Hence its inverse function has range $ℝ$ and domain $\left(1,\infty \right)$. The formula for the inverse is found by rearranging $y={e}^{2x}+1$ to make $x$ the subject. This gives $x=\frac{1}{2}ln\left(y-1\right).$ Hence ${\left(f\circ g\right)}^{-1}\left(x\right)=\frac{1}{2}ln\left(x-1\right).$
Choice (d) is incorrect
Choice (e) is incorrect
Consider the function $f\left(x\right)=\sqrt{3+4x}$ on its natural domain. Find the inverse function ${f}^{-1}$, giving its domain and range. Exactly one option must be correct)
 a) ${f}^{-1}\left(x\right)=\frac{{x}^{2}-3}{4},\phantom{\rule{1em}{0ex}}$ domain $ℝ,$ range $ℝ$ b) ${f}^{-1}\left(x\right)=\frac{1}{\sqrt{3+4x}},\phantom{\rule{1em}{0ex}}$ domain $ℝ,$ range $ℝ$ c) ${f}^{-1}\left(x\right)=\frac{{x}^{2}-3}{4},\phantom{\rule{1em}{0ex}}$ domain $\left[0,\infty \right)$, range $\left[-3∕4,\infty \right)$ d) ${f}^{-1}\left(x\right)=\frac{{x}^{2}-3}{4},\phantom{\rule{1em}{0ex}}$ domain $ℝ$, range $\left[-3∕4,\infty \right)$ e) ${f}^{-1}\left(x\right)=\frac{{x}^{2}-3}{4},\phantom{\rule{1em}{0ex}}$ domain $\left[0,\infty \right)$, range $ℝ$ f) ${f}^{-1}\left(x\right)=\frac{1}{\sqrt{3+4x}},\phantom{\rule{1em}{0ex}}$ domain $\left[0,\infty \right)$, range $ℝ$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The natural domain of $f$ is $\left[-3∕4,\infty \right)$ and the range is $\left[0,\infty \right)$. These are swapped for the inverse function.
Choice (d) is incorrect
Choice (e) is incorrect
Choice (f) is incorrect
What is $\underset{x\to 0}{lim}\frac{sin\left(3x\right)}{x}$ ? Exactly one option must be correct)
 a) $0$ b) $\frac{1}{3}$ c) $1$ d) $3$ e) $\infty$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Make the substitution $t=3x$, so that the limit becomes
$\underset{x\to 0}{lim}\frac{sin\left(3x\right)}{x}=\underset{t\to 0}{lim}\frac{sin\left(t\right)}{\frac{t}{3}}=3\underset{t\to 0}{lim}\frac{sin\left(t\right)}{t}=3.$
Choice (e) is incorrect
What is $\underset{x\to \infty }{lim}\frac{2{x}^{3}-5x+2}{{x}^{3}}$ ? Exactly one option must be correct)
 a) $0$ b) $1$ c) $2$ d) The limit does not exist. e) None of the above

Choice (a) is incorrect
Hint: divide top and bottom by ${x}^{3}$.
Choice (b) is incorrect
Hint: divide top and bottom by ${x}^{3}$.
Choice (c) is correct!
Choice (d) is incorrect
Hint: divide top and bottom by ${x}^{3}$.
Choice (e) is incorrect
Hint: divide top and bottom by ${x}^{3}$.
Determine $\underset{x\to \infty }{lim}\left(\rightln\left(x+1\right)-ln\left(x\right)\left)\right$. Type your answer into the box.

Correct!
$ln\left(x+1\right)-ln\left(x\right)=ln\left(\frac{x+1}{x}\right)=ln\left(1+\frac{1}{x}\right)$
and $\underset{x\to \infty }{lim}\left(1+\frac{1}{x}\right)=ln1=0$.

Recall that $ln\left(a\right)-ln\left(b\right)=ln\frac{a}{b}$.
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,

Find the value of the limit $\underset{x\to 0}{lim}\frac{cosx-1}{x}.$ Type your answer into the box.
$\underset{x\to 0}{lim}\frac{cosx-1}{x}=\underset{x\to 0}{lim}\frac{-sinx}{1}=0$