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

(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 .

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  • $\begingroup$ This was asked before: math.stackexchange.com/questions/174190/… $\endgroup$ Aug 21, 2014 at 14:44
  • $\begingroup$ In fact now I have noticed that you have asked a question about the same limit before: math.stackexchange.com/questions/901018/… $\endgroup$ Aug 21, 2014 at 14:48
  • $\begingroup$ 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 $\endgroup$
    – ys wong
    Aug 21, 2014 at 15:11
  • $\begingroup$ 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. $\endgroup$
    – ys wong
    Aug 21, 2014 at 15:19

5 Answers 5

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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.

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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 $m$

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  • $\begingroup$ this does not answer the question, does it? $\endgroup$
    – Ant
    Aug 21, 2014 at 13:08
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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.

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  • $\begingroup$ @yswong welcome $\endgroup$
    – k99731
    Aug 21, 2014 at 12:21
  • $\begingroup$ For part (a), the method of y=mx can only prove discounity, it cannot be used to prove that the limit exists? $\endgroup$
    – ys wong
    Aug 21, 2014 at 13:03
  • $\begingroup$ @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 $\endgroup$
    – k99731
    Aug 21, 2014 at 13:08
  • $\begingroup$ 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? $\endgroup$
    – ys wong
    Aug 21, 2014 at 13:15
  • $\begingroup$ i dont understand what you are asking. $\endgroup$
    – k99731
    Aug 21, 2014 at 13:43
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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 $(x,y)\neq(0,0)$.

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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.

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