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