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

Question

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 1

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 $-\frac{5\pi}{2}$.

Answer 2

hint: Use $\dfrac{\sin\left(\frac{\pi(x+2)}{2}\right)}{\frac{\pi(x+2)}{2}} \to 1$ as $x \to -2$