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How do I show that this limit exist/does not exist? My assumption is that it does not; however I don't see how I can apply the two-path test to verify this.

$$\lim_{(x,y)\to(0,0)}\dfrac{y+\sin3x-3x}{y+x^5} = L,$$

$$\lim_{y\to0}\dfrac{y+\sin(0)-(0)}{y+(0)^5} = 1,$$

But I don't see any second path that simplifies this well;

$$\lim_{x\to0}\dfrac{(0)+\sin x-x}{(0)+x^5} = \:?$$

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    $\begingroup$ The denominator can be zero for points arbitrarily near to $(0,0)$. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 29 '16 at 10:27
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Hint

Let $f(x,y)=\frac{y+\sin(3x)-3x}{y+x^5}$.

What do you think about $$\lim_{t\to 0}f(0,t)\quad \text{and}\quad \lim_{t\to 0}f(t,0)\ \ ?$$

Edited

$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+o(x^5)$$ therefore $$\frac{\sin(x)-x}{x^5}=\frac{-\frac{x^3}{3!}+\frac{x^5}{5!}+o(x^5)}{x^5}\underset{x\to 0}{\longrightarrow }-\infty .$$

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  • $\begingroup$ Am I allowed to use L'Hôpital on the second one? $\endgroup$ – Frank Vel Jan 29 '16 at 10:22
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    $\begingroup$ It's indeed a possibility. $\endgroup$ – Surb Jan 29 '16 at 10:23

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