Wrong Wolfram Alpha result for $\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^4+x^2+y^4}$?

Question

I'm trying to solve this limit: $$ \lim_{(x,y)\to(0,0)}\frac{xy^4}{x^4+x^2+y^4} $$

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Here's my attempt:

$$0 \le |\frac{xy^4}{x^4+x^2+y^4} - 0| = \frac{|x|y^4}{x^4+x^2+y^4},$$ and since $x^4+x^2 \ge0$ then $\frac{y^4}{x^4+x^2+y^4} \le 1$ so $$ \frac{|x|y^4}{x^4+x^2+y^4} \le |x|,$$ so $$ 0 \le \lim_{(x,y)\to(0,0)}|\frac{xy^4}{x^4+x^2+y^4} - 0| \le \lim_{(x,y)\to(0,0)} |x| = 0, $$ and using the squeeze theorem the limit is $0$.

But if I input the limit in wolfram alpha, it says that the limit doesn't exist. Here is the link to the limit in Wolfram Alpha.

Answer 1

Both of

$$\frac{xy^{4}}{x^{4}+x^{2}+y^{4}},\,\,\frac{xy^{4}}{x^{4}+x^{2}+y^{2}}$$

have limit $0.$ For the first one, you gave a valid proof. The second one follows from this one since the denominator is at least as big, while the numerator is the same.

Answer 2

For \begin{align*} \lim_{(x,y)\rightarrow(0,0)}\dfrac{xy^{4}}{x^{4}+x^{2}+y^{2}}, \end{align*} one does the following step which is similar to your technique: \begin{align*} \left|\dfrac{xy^{4}}{x^{4}+x^{2}+y^{2}}\right|\leq|xy^{2}|\cdot\dfrac{y^{2}}{x^{4}+x^{2}+y^{2}}\leq|x|\cdot y^{2}\rightarrow 0 \end{align*} as $(x,y)\rightarrow(0,0)$.

Answer 3

You can also use polar coordinate by letting $x = r\cos \theta$ and $y = r\sin\theta$ and the limit becomes $$\lim_{r\to 0} \frac{r^5 \cos \theta \sin^4\theta}{r^4\cos^4\theta + r^2\cos^2\theta + r^4\sin^4\theta} = \lim_{r\to 0} r^3\left(\frac{\cos\theta\sin^4\theta}{r^2\cos^4\theta+\cos^2\theta + r^2\sin^4\theta}\right) = 0$$

Today's lesson is : Don't trust Wolframalpha entirely