I found this possible solution:

Let $r^2=x^2 + y^2, x = r \cos(\theta)$ and $y=r\sin(\theta)$. Then:

$$ \begin{split} \lim_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}} &= \lim_{r \to 0} \frac{(r \cos(\theta))^2}{\sqrt{r^2}} \\ &= \lim_{r \to 0} \frac{r^2 \cos^2(\theta)}{r} \\ &= \lim_{r \to 0} r \cos^2(\theta) \end{split}$$

Since $\lvert \cos^2(\theta)\rvert \leq 1$ and $\lim\limits_{r \to 0} r = 0$, we have:

$$ \lim_{r \to 0} r \cos^2(\theta) = 0 = \lim_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}} $$

I learned this substitution in this class.

First of all I would like to know if my solution is possible, but I would also would like to see a prove that does not use this kind of substitution.


3 Answers 3


We have that


Letting $(x,y)\to(0,0)$ we get the desired result.

Also yes your solution works, albeit it's a bit overkill to do a substitution here.


That approach is fine. You can also use the fact that $$ \left|\frac{x^2}{\sqrt{x^2+y^2}}\right|=|x|\left|\frac x{\sqrt{x^2+y^2}}\right|\le|x|, $$ and so, since $\lim_{(x,y)\to(0,0)}|x|=0$, then $\lim_{(x,y)\to(0,0)}\frac{x^2}{\sqrt{x^2+y^2}}=0$.

  • $\begingroup$ I did not understante why: $ |x|\left|\frac x{\sqrt{x^2+y^2}}\right|\le|x| $ $\endgroup$ May 13, 2022 at 21:44
  • 3
    $\begingroup$ Because$$\left|\frac x{\sqrt{x^2+y^2}}\right|=\frac{|x|}{\sqrt{x^2+y^2}}=\frac{\sqrt{x^2}}{\sqrt{x^2+y^2}}=\sqrt{\frac{x^2}{x^2+y^2}}\le1.$$ $\endgroup$ May 13, 2022 at 21:45

You can consider that the following generalized limit:

$$\lim_{(x,y) \to (0,r)} \frac{x^2}{\sqrt{x^2 +y^2}}=0.$$

Observe that,

$$\begin{align}\frac{x^2}{\sqrt{x^2}}&=\frac{x^2}{|x|}=|x|\\ &≥\frac{x^2}{\sqrt{x^2 +y^2}}\\ &≥0\end{align}$$

This implies

$$\begin{align}0&=\lim_{(x,y) \to (0,0)}|x|\\ &=\lim_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}} \end{align}$$

This also shows that

$$\begin{align}\lim_{(x,y) \to (0,0)}|x|&=\lim_{(x,y) \to (0,r)} \frac{x^2}{\sqrt{x^2 +y^2}}&\\ &=0. \end{align}$$

where, $r\in\mathbb R$.

  • $\begingroup$ Notice that if $r\neq0$ the limit is trivial by continuity $\endgroup$
    – Lorago
    May 13, 2022 at 23:04

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