<s>[INST] Limit for $(x,y) \to (0,0)$ of $f(x,y)=\frac{x^2+y^4}{x}$ I tried to solve this limit, but I'm not sure whether it's correct. The candidate limit is $l=0$.
From $ | \frac{x^2+y^4}{x} | $, since $\sqrt{x^2-y^2} \leq \sqrt{x^2} = |x| $, we then have:
$$| \frac{x^2+y^4}{x} | \leq | \frac{x^2+y^4}{\sqrt{x^2-y^2}} | \leq ...$$
$$ ...\leq | \frac{r^2 cos^2 {\theta}+r^4 sin^4 {\theta}}{\sqrt{r^2 (cos^2 {\theta} - sin^2 {\theta})}} | \leq | \frac{r cos^2 {\theta}+r^3 sin^4 {\theta}}{\sqrt{(cos^2 {\theta} - sin^2 {\theta})}} \leq ... $$
$$ ... \leq | \frac{r+r^3}{\sqrt{cos(2\theta)}} | $$
The only case where the denominator is 0 is when $\theta = \frac{\pi}{2}$. However, this restriction is not a subset of the domain because we have $x \neq 0$ by hypothesis.
So the limit is 0.
Is this correct? I am not entirely sure about the last part!
 [/INST] I suggest consider curve $y=\sqrt[8]{x}=x^{\frac{1}{8}}$. Then we have on this curve
$$\frac{x^2+y^4}{x}=x + \frac{1}{\sqrt{x}}$$
limit of first summand exists, while second tends to infinity.
On another hand (accordingly comments) along the curve $y=x$ limit is obviously zero. Thus, we can postulate, that limit does not exist.
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