Compute, without L'Hopital's rule, the limit $$\lim_{x\to -2^-}\frac{\sin(x+2)}{|4-x^2|}$$
Since $x\to -2^-$ , the denominator can be rewritten as $-4+x^2$, but there isn't much more I've been able to do (I tried using $\sin(x+2)=\sin x\cos2+\sin2\cos x$ without getting much out of it). Thanks for your answers.
$$\lim_{x\to -2^-}\frac{\sin(x+2)}{|4-x^2|}=\lim_{x\to -2^-}\frac{\sin(x+2)}{x^2-4}=\lim_{x\to -2^-}\frac{\sin(x+2)}{(x+2)(x-2)}=\lim_{u\to 0^-}\frac{\sin(u)}{u(u-4)}=...$$