Answer
The limit is
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{x^2+y^2}=0.$$
Work Step by Step
We have
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{x^2+y^2}.$$
Introduce the polar coordinates:
$$\rho^2=x^2+y^2,\quad \tan\frac{y}{x}, \quad (x,y)\to0 \Rightarrow \rho\to0.$$
The limit becomes
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{x^2+y^2}=\lim_{\rho\to0}\frac{1-\cos\rho^2}{\rho^2}.$$
When $\rho\to0$ then $1-\cos\rho^2 \to 0$ because $\cos\rho^2\to 1$. Also in the denominator $\rho^2\to0$. Since both the numerator and the denominator tend to zero we can use L'Hopital's rule:
$$\lim_{\rho\to0}\frac{1-\cos\rho^2}{\rho^2}=\lim_{\rho\to0}\frac{(1-\cos\rho^2)'}{(\rho^2)'}=\\ \lim_{\rho\to0}\frac{2\rho\sin\rho}{2\rho}=\lim_{\rho\to0}\sin\rho=0$$
where in the last step we used the metod of substitution to evaluate the limit.
