## MATH1901 Quizzes

Quiz 5: Limits and the limit laws
Question 1 Questions
What is the value of the limit $\underset{x\to 1}{lim}\frac{{x}^{2}-x-2}{{x}^{2}-2x}$ ? Exactly one option must be correct)
 a) $-2$ b) $-1$ c) $1$ d) $2$ e) This limit does not exist.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$\underset{x\to 1}{lim}\frac{{x}^{2}-x-2}{{x}^{2}-2x}=\frac{1-1-2}{1-2}=2.$
Choice (e) is incorrect
What is value of the limit $\underset{x\to 0}{lim}\frac{{x}^{2}-x-2}{{x}^{2}-2x}$ ? Exactly one option must be correct)
 a) $-2$ b) $1$ c) The limit does not exist d) $\infty$ e) $-\infty$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
We can reduce this limit to the limit $\underset{x\to 0}{lim}\frac{1}{x}$, which we know does not exist: $\begin{array}{llll}\hfill \underset{x\to 0}{lim}\phantom{\rule{2.77695pt}{0ex}}\frac{{x}^{2}-x-2}{{x}^{2}-2x}& =\underset{x\to 0}{lim}\phantom{\rule{2.77695pt}{0ex}}\frac{{x}^{2}-2x+x-2}{{x}^{2}-2x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\underset{x\to 0}{lim}\phantom{\rule{2.77695pt}{0ex}}1+\frac{x-2}{x\left(x-2\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1+\underset{x\to 0}{lim}\phantom{\rule{2.77695pt}{0ex}}\frac{1}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (d) is incorrect
Choice (e) is incorrect
What is $\underset{x\to 2}{lim}\frac{{x}^{2}-x-2}{{x}^{2}-2x}$ ? Exactly one option must be correct)
 a) $0$ b) $1$ c) $\frac{3}{2}$ d) The limit does not exist. e) $\infty$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
As in the last question, $\begin{array}{llll}\hfill \underset{x\to 2}{lim}\frac{{x}^{2}-x-2}{{x}^{2}-2x}& =\underset{x\to 2}{lim}\frac{{x}^{2}-2x+x-2}{{x}^{2}-2x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\underset{x\to 2}{lim}1+\frac{x-2}{x\left(x-2\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1+\underset{x\to 2}{lim}\frac{1}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1+\frac{1}{2}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (d) is incorrect
Choice (e) is incorrect
What is $\underset{x\to 1}{lim}\sqrt{1+3x}-1$ ? Type your answer into the box.

Correct!
$\underset{x\to 1}{lim}\sqrt{1+3x}-1=\sqrt{1+3}-1=1.$
Try substituting $x=1$ into $\sqrt{1+3x}-1$.
What is $\underset{x\to 0}{lim}\frac{sin\left(3x\right)}{x}$ ? Type your answer into the box. 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) $\infty$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Choice (e) is incorrect
Use the squeeze law to find $\underset{x\to 0}{lim}5{x}^{2}\left(1-cosx\right)$. Type your answer into the box.

Correct!
Since $0\le 1-cosx\le 2$ we see that $0\le 5{x}^{2}\left(1-cosx\right)\le 10{x}^{2}$. Therefore, by the squeeze law
$0\le \underset{x\to 0}{lim}5{x}^{2}\left(1-cosx\right)\le \underset{x\to 0}{lim}10{x}^{2}=0.$

Try using the inequality $-1\le cosx\le 1$.
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 $ln\left(a\right)-ln\left(b\right)=ln\frac{a}{b}$.
What is $\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-2{y}^{2}}{3{x}^{2}+{y}^{4}}$ as $\left(x,y\right)\to \left(0,0\right)$ along the $x-$axis? Exactly one option must be correct)
 a) $0$ b) $\frac{1}{3}$ c) $1$ d) $\infty$

Choice (a) is incorrect
Choice (b) is correct!
Along the $x$–axis we have $y=0$ so this limit becomes

$\underset{x\to 0}{lim}\frac{{x}^{2}}{3{x}^{2}}=\underset{x\to 0}{lim}\frac{1}{3}=\frac{1}{3}.$
Choice (c) is incorrect
Choice (d) is incorrect
What is the value of the limit $\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-2{y}^{2}}{3{x}^{2}+{y}^{4}}$ if $\left(x,y\right)$ approaches $\left(0,0\right)$ along the line $y=x$? Exactly one option must be correct)
 a) $-\frac{2}{3}$ b) $-\frac{1}{3}$ c) $0$ d) $\infty$

Choice (a) is incorrect
Choice (b) is correct!
Along the line $y=x$ this limit becomes
$\underset{x\to 0}{lim}\frac{{x}^{2}-2{x}^{2}}{3{x}^{2}+{x}^{4}}=\underset{x\to 0}{lim}\frac{-{x}^{2}}{3{x}^{2}+{x}^{4}}=\underset{x\to 0}{lim}\frac{-1}{3+{x}^{2}}=-\frac{1}{3}.$
Notice that Question 9 combined with Question 10 shows that the limit $\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-2{y}^{2}}{3{x}^{2}+{y}^{4}}$ does not exist.
Choice (c) is incorrect
Choice (d) is incorrect