Find $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$ $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$
I tried replacing x and y with several values and kept getting 1 so I tried:
$$0 \le |\cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})| \le |\cos(|(\frac{x^2-y^2}{\sqrt{x^2+y^2}}|)|\le |\cos(\frac{|x|^2+|y|^2}{\sqrt{x^2+y^2}})|\le |\cos(\frac{2\sqrt{x^2+y^2}^2}{\sqrt{x^2+y^2}})| \le |\cos(2\sqrt{x^2+y^2})|$$
Is this correct? I couldn't get Wolfram to compute this properly for some reason.
 A: The hint:
Prove that $$-(|x|+|y|)\leq\frac{x^2-y^2}{\sqrt{x^2+y^2}}\leq|x|+|y|.$$
The right inequality.
We need to prove that:
$$\frac{(|x|-|y|)(|x|+|y|)}{\sqrt{x^2+y^2}}\leq|x|+|y|$$ or $$ |x|-|y|\leq\sqrt{x^2+y^2}.$$
If $|x|-|y|\leq0$ it's obvious.
But for $|x|-|y|\geq0$ it's enough to prove that
$$(|x|-|y|)^2\leq x^2+y^2$$ or
$$-2|xy|\leq0,$$ which is obvious. 
By the same way we can prove a left inequality. 
Thus, since $$\lim_{(x,y)\rightarrow(0,0)}(|x|+|y|)=\lim_{(x,y)\rightarrow(0,0)}(-(|x|+|y|))=0,$$
we obtain: $$\lim_{(x,y)\rightarrow(0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}=0.$$
Id est, since $\cos$ is a continuous function, we obtain:
$$\lim_{(x,y)\rightarrow(0,0)}\cos\frac{x^2-y^2}{\sqrt{x^2+y^2}}=\cos\left(\lim_{(x,y)\rightarrow(0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}\right)=\cos0=1.$$
A: HINT
Consider the polar coordinates $x = r\cos(\theta)$ and $y = r\sin(\theta)$. Then we get
\begin{align*}
\lim_{(x,y)\rightarrow(0,0)}\cos\left(\frac{x^{2}-y^{2}}{\sqrt{x^{2}+y^{2}}}\right) = \lim_{r\rightarrow 0}\cos(r\cos(2\theta)) = \ldots
\end{align*}
Can you take it from here?
A: You can do directly about the $x,y$ :  $ 0 \le \dfrac{x^2}{\sqrt{x^2+y^2}} \le |x|$ and similarly for the other one. This means ...the expression in the $\cos \to 0$ and the answer is clearly $1$.
