Is this correct? Does no limit exist?
Solution 1:
Again we apply L'Hospitals Rule:
$$\frac{1}{2}\cdot\lim_{x\to0} \frac{8x\cdot\cos(4x^2)}{\color{red}2x} = \frac{1}{2}\cdot\lim_{x\to0}\color{red}4\cdot\cos(4x^2) =\color{red}2 \lim_{x\to0} \cos(4x^2) = \color{red}2 $$
Solution 2:
I think the best approach is to just make the substitution $u = x^2 + y^2$.
Then $g$ is continuous at $(0,0)$ if and only if $$\lim_{u\to0^{+}}\frac{\sin(2u)}{u}=a$$ but then, since $$\lim_{u\to0^{+}}\frac{\sin(2u)}{u}=2\left(\lim_{u\to0^{+}}\frac{\sin(2u)}{2u}\right)=2(1)=2$$ we get continuity at $(0,0)$ if and only if $a = 2$.