calculus - Finding $lim_{xto -2}~~ sin(frac{pi x}{2})frac{x^2+1}{x+2}$.

I would really appreciate if you could help me solving this limit problem!

Find the limit without using L'Hopital's rule!

$$\lim_{x\to -2} \sin\bigg(\frac{\pi x}{2}\bigg)\frac{x^2+1}{x+2} = ?$$

Thank you in advance!

Answer

First, pull out $\lim_{x\to -2} (x^2+1)$ from the limit. Then, set $y=x+2$, which gives

$$\lim_{y\to 0}\frac{\sin \big((\frac{\pi}{2}y)-(\frac{\pi}{2}2)\big)}{y}=$$

$$=\lim_{y\to 0}\frac{-\sin(\frac{\pi}{2}y)}{y}=$$

$$=\lim_{y\to 0}\frac{-\frac{\pi}{2}\sin(\frac{\pi}{2}y)}{\frac{\pi}{2}y}=$$

$$=-\frac{\pi}{2}\lim_{\frac{\pi}{2}y\to 0}\frac{\sin(\frac{\pi}{2}y)}{\frac{\pi}{2}y}=$$
$$=-\frac{\pi}{2}$$

Now combine with the $5$ you pulled out previously to get $-\frac{5\pi}{2}$.