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How to compute ? $$\lim_{(x,y) \rightarrow (0,0)} \frac{xy^4}{x^2+y^8}$$ I feel the limit is $0$ buy I'm uncertain how to show that.

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    $\begingroup$ That's a classic counterexample: the limit along every straight line is 0, but the overall limit doesn't exist. $\endgroup$ Sep 29, 2016 at 12:49

2 Answers 2

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Setting $y = \sqrt{x}$, we obtain:

$\displaystyle \lim_{x \to 0} \frac{x^3}{x^2+x^4} = \frac{x}{1+x^2} = 0$

Setting $y = \sqrt[4]{x}$, we obtain:

$\displaystyle \lim_{x \to 0} \frac{x^2}{x^2+x^2} = \frac{1}{2}$

For the limit to exist, it can only take on a unique value. Thus, the limit does not exist.

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we have $f(y^4,y)=\frac{1}{2}$

and $f(2y^4,y)=\frac{2}{17}$

so when $y$ tends to zero, we have two values. the limit doesn't exist.

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