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Question 5 (text reproduced below)

The function $f\colon\mathbb{R}^2\to\mathbb{R}$ is given by $$ f(x,y) = \frac{(x-1)(y+1)^3}{(x-1)^2+(y+1)^6} $$ a. Approach the point $(1,-1)$ by straight lines and calculate the limit of $f(x,y)$ along those straight-line paths. (Hint: $\lim_{u\to0} f(1+u, -1+ku)$ for $\in\mathbb{R}$.)

b. Show that there is another path along which $f(x,y)$ limit as calculated in part (a), as $(x,y)$ approaches $(1,-1)$. Does the function $f$ have a limit as $(x,y)$ tends to $(1,-1)$?

Question B asks to show another path with a different limit for $f(x,y)$. Yet the limit exist what other path can approach the function without contradicting with the existence of the limit.

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  • $\begingroup$ If you are asked to find two paths yielding different limits, why would you think the function has a limit overall? $\endgroup$ – Clement C. Aug 11 '16 at 11:39
  • $\begingroup$ Because the limit does exist wolframalpha.com/input/…. But it looks the question is contradicting itself or I missing something huge. $\endgroup$ – encryped Aug 11 '16 at 11:41
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    $\begingroup$ You may be missing something in that you have an utter faith in Wolfram Alpha. $\endgroup$ – Clement C. Aug 11 '16 at 11:47
  • $\begingroup$ Just as a fun fact: as right now wolframalpha also shows that the limit does not exists. $\endgroup$ – hal4math Sep 17 at 21:25
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The limit does not exist, even if you seem to think so based on Wolfram Alpha.

First advice: in these situations, do not take Wolfram Alpha's result as the word of the law.

Now, try the parameterization $(x,y)=(1+u^3,-1+u)$, when $u\to 0$. Spoiler below: place your mouse on it to reveal the text.

$$f(x,y) = \frac{(x-1)(y+1)^3}{(x-1)^2+(y+1)^6}= \frac{u^3\cdot u^3}{u^6+u^6} = \frac{1}{2}\xrightarrow[u\to0]{} \frac{1}{2}.$$

The idea? We want to chose $a,b>0$ such that $x=1+u^a$, $y=-1+u^b$ lead to nice simplifications in $f(x,y)$, with a limit that is not $0$. The exponents above do that.

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  • $\begingroup$ Thank you for you explaination, I will no longer blindly rely on Wolfram Alpha. $\endgroup$ – encryped Aug 11 '16 at 12:47
  • $\begingroup$ In my experience (limited, I have to say), Mathematica/Wolfram is good for univariate limits, but has to be taken with a grain of salt (or, rather, give results hard to interpret) in multivariate or more complicated scenarii. $\endgroup$ – Clement C. Aug 11 '16 at 12:51

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