Limit with 2 variables: $\frac{x^2y}{x^4 +y^2}$, $\frac{e^{xy^3}-1}{x^2 +y^4}$ and $(x^2 +y^2)^{xy}$ at $(0,0)$

Question

I am new to this topic so I appreciate any help on this.

a)$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{x^2y}{x^4 +y^2}$$

For $x=0 \lor y=0 : f(x,y) \longrightarrow 0$ but for $y = x^2$ the limit is $\frac{1}{2}$ so the limit does not exist.

b) $$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{e^{xy^3}-1}{x^2 +y^4}$$ I don't have any ideas here

c) $$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = (x^2 +y^2)^{xy}$$ Let $x \geq y$ then $|x^2 +y^2|^{xy} \leq |2y^2|^{y^2} \longrightarrow 1 \quad $for $y \longrightarrow 0$ but for $x = \frac{1}{y} \Longrightarrow f(x,y) = (\frac{1}{y^2} + y^2)^1 \longrightarrow \infty \quad$ for $y \longrightarrow 0 $ so the limit does not exist.

Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. — Martin Sleziak, May 21 '17 at 23:42
The first limit seems to be posted here rather frequently: http://math.stackexchange.com/questions/933949/multivariable-limit-frac2x2yx4y2 http://math.stackexchange.com/questions/1476353/help-with-lim-x-y-to0-0-fracx2yx4y2 http://math.stackexchange.com/questions/93886/computing-a-multivariable-limit http://math.stackexchange.com/questions/174190/prove-that-lim-limits-x-y-to-0-0-fracxy2x2-y4-0 http://math.stackexchange.com/questions/1780919/showing-a-function-is-discontinuous-by-taking-a-limit http://math.stackexchange.com/questions/1837256 — Martin Sleziak, May 21 '17 at 23:43
Using Approach0 I found this question about the limit c: Multivariable limit $\lim_{{x, y \to 0}}(x^2+y^2)^{x^2y^2}=1$ — Martin Sleziak, May 21 '17 at 23:51

score 5 · Accepted Answer · answered May 21 '17 at 16:48

5

For b):

we have that $f(0,y) = f(x, 0) = 0$ for every $x,y\neq 0$. If $x$ and $y$ are both non zero, we have that $$ \frac{e^{xy^3} - 1}{x^2+y^4} = \frac{e^{xy^3} - 1}{xy^3}\cdot \frac{xy^3}{x^2+y^4}. $$ The first factor at the r.h.s. is bounded (for $x,y$ bounded), since $(e^t-1)/t \to 1$ as $t\to 0$.

The second factor can be estimated by $$ \left|\frac{xy^3}{x^2+y^4}\right| = |y| \cdot \frac{|x| y^2}{x^2+y^4}\leq \frac{|y|}{2}. $$

Hence the limit exists and is equal to $0$.

answered May 21 '17 at 16:48

Rigel

14,434

1

I see, i could have used a) for this. Thats very helpful – May 21 '17 at 17:03
what makes the second factor bounded by y/2? we cannot say the max of second factor is 1/2 through part a – james black Sep 20 '19 at 03:25
$|ab| / (a^2+b^2) \leq 1/2$ for every $(a,b)\neq (0,0)$. – Rigel Sep 24 '19 at 06:41

score 2 · Answer 2 · answered May 21 '17 at 18:03

Another way to process is to set $y=mx^\alpha$ with some well chosen $\alpha>0$ so that the various sums $x^a+y^b$ become homogeneous.

Of course $m=m(x,y)$ is not a constant, just the ratio of the rate of convergence of $x$ and $y$.

Part a:

With $y=mx^2$

$\displaystyle f(x,y)=\frac{x^2y}{x^4+y^2}=\frac{mx^4}{x^4+m^2x^4}=\frac{m}{1+m^2}$

This is bounded but can take multiple values (e.g $m=1,\ f(x,x^2)=\frac 12$ and $m=0,\ f(x,0)=0$), so there is no limit in $(0,0)$.

Part b:

With $x=my^2$

$\displaystyle f(x,y)=\frac{e^{x^3y}-1}{x^2+y^4}\sim\frac {x^3y}{x^2+y^4}=\frac{m^3y^7}{m^2y^4+y^4}=\frac{m^3y^3}{1+m^2}=\underbrace{\bigg(\frac{m^2}{1+m^2}\bigg)}_\text{bounded}xy\to 0$

So there is a limit and it is $0$.

Part c:

The previous method do not work well here, but polar coordinates do

$f(x,y)=(x^2+y^2)^{xy}=\exp(xy\ln(x^2+y^2))=\exp(\underbrace{\cos(\theta)\sin(\theta)}_\text{bounded}\underbrace{r^2\ln(r^2)}_{\to 0})\to e^0=1$

Be careful. Looking along families of curves will only prove a limit does not exist, not that it does. — Ted Shifrin, May 22 '17 at 15:58
Please read, m is not a constant, no matter how many time I write this, people don't see it. These are not curves. — zwim, May 22 '17 at 16:12
It's still very misleading. How do you handle $f(x,y) = \begin{cases} 1, & |y|>|x|^3 \text{ or } y=0 \ 0, & \text{otherwise} \end{cases}$ or $f(x,y) = \begin{cases} 1, & |y|>e^{-1/x^2} \text{ or } y=0 \ 0, & \text{otherwise}\end{cases}$? — Ted Shifrin, May 22 '17 at 16:18

zhw. · Answer 3 · 2017-05-21T17:10:08.763

1

You've made a mistake in c): $(1/y,y)$ does not approach $(0,0)$ as $y\to 0.$

To do it, we can apply $\ln$ to get $xy\ln (x^2+y^2).$ Recall $|xy|\le (x^2+y^2)/2.$ Thus

$$\tag 1 |xy \ln (x^2+y^2)| \le \frac{(x^2+y^2)\ln (x^2+y^2)}{2}.$$

Now $u\ln u \to 0$ as $u\to 0^+.$ Thus the limit of $(1)$ as $(x,y) \to (0,0)$ is $0.$ This implies the orginal limit is $1.$

edited May 21 '17 at 17:10

answered May 21 '17 at 17:02

zhw.

105,693

Limit with 2 variables: $\frac{x^2y}{x^4 +y^2}$, $\frac{e^{xy^3}-1}{x^2 +y^4}$ and $(x^2 +y^2)^{xy}$ at $(0,0)$

3 Answers3

Part a:

Part b:

Part c: