# Multivariate limit $\lim_{(x,y) \to (0,0)} \frac{{x{y^2}}}{{{x^2} + {y^4}}} = 0$

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

(a) Prove that the limit of $f(x, y)$ as $(x, y)$ approaches $(0, 0)$ along any straight line is $0$.

(b) Does $\lim_{(x,y)\to(0,0)} f(x, y)$ exist?

What I'm confused about this question is, for part (b) based on the discounity test the limit clearly does not exist. If we let $x=y^2$ which gives a limit of $0.5$ and if we let $x=y$, the limit approaches $0$. But in part (a), how can the limit approach $0$ when it does not even exist? And another point is that for part (a), we cannot let y=mx to prove that the limit exists along a straight line because that method can only test for discounity, it cannot be used to prove that a limit exists? Note: what this question is asking is that even though the limits clearly does not exists, we have to prove why it does seem to exists at 0 when we ONLY consider the approach path of the straight .

• This was asked before: math.stackexchange.com/questions/174190/… Aug 21, 2014 at 14:44
• In fact now I have noticed that you have asked a question about the same limit before: math.stackexchange.com/questions/901018/… Aug 21, 2014 at 14:48
• Yeah this meant as an add on to that question and it is a little different because the question compares finding limits via a straight line and finding limits through a curve which gets confusing. Because the limits actually do exists when ONLY consider the straight line but do NOT exists when we approach the limits through curves Aug 21, 2014 at 15:11
• Note: what this question is asking is that even though the limits clearly does not exists, we have to prove why it does seem to exists at 0 when we ONLY consider the approach path of the straight line. Aug 21, 2014 at 15:19

In part (a) you only have to show that the limit approaches to 0 if you move along a straight line. This is not a contradiction to your result for $x = y^2$ as $(x, y)$ then moves on a parabola.

Choose path $x=my^2$ then $\lim_{(x,y)\to (0,0)} f(x,y)={m\over 1+m^2}$ which is different for different $m$

• this does not answer the question, does it?
– Ant
Aug 21, 2014 at 13:08

Part (a) emphasise straight line. So we can cosider paths $y=mx$ and $x=0$.

Along $y=mx$, we have $f= \frac{m^2x^3}{x^2+m^4x^4}$

Eliminating $x^2$, we have $\frac{m^2x}{1+m^4x^2}$, which goes to 0 as x goes to 0.

Along $x=0$, we have it 0. So limit of 0 is 0.

• @yswong welcome Aug 21, 2014 at 12:21
• For part (a), the method of y=mx can only prove discounity, it cannot be used to prove that the limit exists? Aug 21, 2014 at 13:03
• @yswong yes. more precisely, you cannot use y=mx to disprove continuity, since it has limit 0. to disprove continuity, you have to take y=mx^2 in this case. by the way, you can use x=rcos(t) and y=rsin(t) to prove continuity Aug 21, 2014 at 13:08
• But again part(a) mentions only on a straight line, does the method of y=mx^2 or polar coords works on a straight line? Aug 21, 2014 at 13:15
• i dont understand what you are asking. Aug 21, 2014 at 13:43

In a) you are actually asked to prove that $$\lim_{\lambda\rightarrow 0}\frac{(\lambda x)(\lambda y)^2}{(\lambda x)^2+(\lambda y)^4}=0$$ for each fixed $(x,y)\neq(0,0)$.

Although the straight lines all pass through 0.1, 0,01, 0.001 to reach zero, they do so at different distances from the origin. So it is possible for the $f(x,y)=0.01$ curve to approach the origin, so long as all straight lines come inside it eventually.