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I've been trying to prove the existence of this limit... trying with a bunch of different curves I'm always getting $0$ (I don't know if I'm using the suggestion correctly) but when I try to make an epsilon-delta proof I get stucked because I don't know how to properly bound the denominator.

$$\lim_{(x,y) \rightarrow (0,2)} \frac{x (y-2)^3}{3x - 5(y-2)^4}$$

(Consider curves close to $3x-5(y-2)^4=0$)

My question is: if indeed this limit does exist -and it's $0$- how can I prove it? As I wrote above I don't know how to work around the denominator. I need something smaller than $|3x-5(y-2)^4|$ ...

Or maybe the limit does not exist, and then my question would be what curve can you think of to prove it.

Thank you for your time!

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Follow the curve $$x = \frac{5Rt^4}{3R-t^3}, \quad y = 2+t$$ where $R \in \mathbb R$. What is the limit?

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  • $\begingroup$ Would you give us some intuition as how do you arrive to that curve? $\endgroup$
    – Miguel
    Jul 14, 2017 at 15:18
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    $\begingroup$ I just solved $\frac{x(y-2)^3}{3x-5(y-2)^4} = R$ for $x$ (which was trivial, while solving for $y$ is difficult). $\endgroup$
    – md2perpe
    Jul 14, 2017 at 15:32

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