# Prove that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}} = 0$ without substituting

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.

We have that

$$0\leq\frac{x^2}{\sqrt{x^2+y^2}}\leq\frac{x^2+y^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}.$$

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

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

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