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

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

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 $$|f(\x)| <\epsilon.$$

• Your first inequality is not right. Replacing $x_1$ and $x_2$ with $\delta$ makes the denominator as large as possible, and hence makes the whole quotient smaller rather than larger as claimed. For instance consider your argument applied to just $x_2^2/x_1$.
– user208649
Apr 14, 2020 at 16:14

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 $$\lim_{x\to0} f(x) = 0.$$
• Ah I'm too used to looking in $\mathbb{C}$, was going to ask how you got $|\frac{x_1}{x_1^4/x_2^2 + 1}| < |x_1|$ :P - anyhow, nicely done +1. Apr 14, 2020 at 17:11
• Minor quibble: The first two strict inequalties in the display aren't really correct if $x_1=0$. Better, I think, might be to note that $|x_2^2/(x_1^4+x_2^2)|\le1$ if $(x_1,x_2)\not=(0,0)$. Apr 23, 2020 at 21:46
Aliter: $$0< \left| \frac{x_1x_2^2}{x_1^4+x_2^2}\right|<|x_1|$$. Taking the limit we get the answer.