Proving that the $\lim_{x\to 0} f(x)=0$

Question

Can I please receive feedback on my proof below? Thank you!

$\def\R{{\mathbb R}} \def\x{{\bf x}} \def\0{{\bf 0}}$

Let $f\colon \R^2\to \R$ be given by $$f(\x)=f(x_1,x_2) = \left\{\begin{array}{cl} \frac{x_1 x^2_2}{x^4_1+x^2_2} & \mbox{if $\x\ne\0$,} \\ 0 & \mbox{if $\x=\0$.} \end{array}\right.$$ Prove that $\displaystyle{\lim_{\x\to\0} f(\x)=0}$.

$\textbf{Solution:}$ Let us consider that $||\x|| <\delta$. Hence, $x_1<\delta$ and $x_2 < \delta$. Now in this situation for $\x \ne \0$ $$f(\x) = \frac{x_1x_2^2}{x_1^4 + x_2^2} < \frac{\delta^3}{\delta^4 + \delta^2} = \frac{\delta}{1+ \delta^2}.$$ Therefore, $\displaystyle{\epsilon=\frac{\delta}{1+\delta^2}}$. Now, $1+\delta^2$ is always positive. Hence $\epsilon >0$ and $\delta >0$. Therefore, for $\epsilon > 0$, we will find $\delta >0$, such that $||\x|| < \delta$ implies $|f(\x)| <\epsilon.$

Answer 1

As pointed out in the comment section your manipulation is not right.

In order to prove continuity at $0$, and therefore your limit. Given $\varepsilon>0$, consider $\delta = \varepsilon$. Therefore, if $\|(x_1,x_2)\|<\delta = \varepsilon$ $$|f(x_1,x_2)|=\left|\frac{x_1x_2^2}{x_1^4 + x_2^2}\right| = \left|\frac{x_1}{x_1^4/x_2^2 + 1}\right| \leq |x_1| <\delta=\varepsilon, \quad \mbox{if $x_2\neq 0$}.$$

Note that if $x_2 = 0$, then $f(x_1,x_2)=0$. Therefore $\|(x_1,x_2)\|<\delta $ $\Rightarrow$ $|f(x_1,x_2)|<\varepsilon$, so $\lim_{x\to0} f(x) = 0.$

Answer 2

Aliter: $0< \left| \frac{x_1x_2^2}{x_1^4+x_2^2}\right|<|x_1|$. Taking the limit we get the answer.