## MATH1001 Quizzes

Quiz 6: Partial derivatives, limits and continuity
Question 1 Questions
The function $f$ of two real variables $x$ and $y$ has the property that ${f}_{y}\left(x,y\right)=3{x}^{2}y-{y}^{3}.$ Check all options which are possible values of $f\left(x,y\right)$. (Zero or more options can be correct)
 a) $3{x}^{2}-3{y}^{2}$ b) $\frac{3}{2}{x}^{2}{y}^{2}-\frac{1}{4}{y}^{4}+lnx$ c) $\frac{3}{2}{x}^{2}{y}^{2}-\frac{1}{4}{y}^{4}+y{e}^{x}$ d) $\frac{3}{2}{x}^{2}{y}^{2}-\frac{1}{4}{y}^{4}+sinx$

There is at least one mistake.
For example, choice (a) should be False.
There is at least one mistake.
For example, choice (b) should be True.
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 True.
Correct!
1. False
2. True
3. False
4. True
For which functions do the second partial derivatives $\frac{{\partial }^{2}f}{\partial {x}^{2}}$ and $\frac{{\partial }^{2}f}{\partial {y}^{2}}$ add together to give zero? Check each option with this property. (Zero or more options can be correct)
 a) $f\left(x,y\right)={x}^{2}y-4x{y}^{3}$ b) $f\left(x,y\right)={x}^{2}-{y}^{2}$ c) $f\left(x,y\right)=xlny$ d) $f\left(x,y\right)=xy+{e}^{x}siny$ e) $f\left(x,y\right)=\sqrt{x+y}$

There is at least one mistake.
For example, choice (a) should be False.
In this case we have ${f}_{xx}=2y$ and ${f}_{yy}=-24xy$.
There is at least one mistake.
For example, choice (b) should be True.
In this case we have ${f}_{xx}=2$ and ${f}_{yy}=-2$, which add to zero.
There is at least one mistake.
For example, choice (c) should be False.
In this case we have ${f}_{xx}=0$ and ${f}_{yy}=-\frac{x}{{y}^{2}}$.
There is at least one mistake.
For example, choice (d) should be True.
In this case we have ${f}_{xx}={e}^{x}siny$ and ${f}_{yy}=-{e}^{x}siny$, which add to zero.
There is at least one mistake.
For example, choice (e) should be False.
In this case we have ${f}_{xx}=-\frac{1}{4}{\left(x+y\right)}^{-3∕2}={f}_{yy}$.
Correct!
1. False In this case we have ${f}_{xx}=2y$ and ${f}_{yy}=-24xy$.
2. True In this case we have ${f}_{xx}=2$ and ${f}_{yy}=-2$, which add to zero.
3. False In this case we have ${f}_{xx}=0$ and ${f}_{yy}=-\frac{x}{{y}^{2}}$.
4. True In this case we have ${f}_{xx}={e}^{x}siny$ and ${f}_{yy}=-{e}^{x}siny$, which add to zero.
5. False In this case we have ${f}_{xx}=-\frac{1}{4}{\left(x+y\right)}^{-3∕2}={f}_{yy}$.
What is the value of $\underset{x\to 1}{lim}\frac{{x}^{2}-x-2}{{x}^{2}-2x}$ ? (Hint: Decide whether the function is a continuous function in an interval about the point $x=1$.) 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!
The function is defined and continuous at $x=1$. Therefore the limit is simply the function value at this point. $\underset{x\to 1}{lim}\frac{{x}^{2}-x-2}{{x}^{2}-2x}=\frac{1-1-2}{1-2}=2.$
Choice (e) is incorrect
What is $\underset{x\to 2}{lim}\frac{{x}^{2}-x-2}{{x}^{2}-2x}$ ? (Hint: Factorise and simplify the expression.) 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!
$\begin{array}{llll}\hfill \underset{x\to 2}{lim}\phantom{\rule{0.3em}{0ex}}\frac{{x}^{2}-x-2}{{x}^{2}-2x}& =\underset{x\to 2}{lim}\phantom{\rule{0.3em}{0ex}}\frac{\left(x-2\right)\left(x+1\right)}{x\left(x-2\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\underset{x\to 2}{lim}\phantom{\rule{0.3em}{0ex}}\frac{x+1}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\underset{x\to 2}{lim}\phantom{\rule{0.3em}{0ex}}1+\frac{1}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1+\frac{1}{2}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
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 $\sqrt{1+3x}-1$. (This is valid because the function is continuous at $x=1$.)
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) Function values increase without bound as $\left(x,y\right)\to \left(0,0\right)$ and so there is no limit.

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) Function values increase without bound as $\left(x,y\right)\to \left(0,0\right)$ and so there is no limit.

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 this question combined with the previous one 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
At what value or values of $x$ is the function
NOT continuous? (You may check more than one box if you wish.) (Zero or more options can be correct)
 a) $-1$ b) 0 c) 1 d) 2 e) No values ($f\left(x\right)$ 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
NOT continuous? Exactly one option must be correct)
 a) $-1$, $0$ and $1$ b) 1 c) 0 d) $0$ and $1$ e) No values ($f\left(x\right)$ 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 $a=-1$.