Evaluate $\lim_{x \to -2} \frac{x + 2}{\sin(\frac{\pi x}{2})}$ using continuity (without L'Hospital) 
A little background: I am TAing for a Calculus I class (mostly non-math majors), and this problem showed up on the worksheet for tomorrow. We have fully developed limits, including the squeeze theorem, and continuity, but we have not covered L'Hospital's rule yet. I am not sure how the students are supposed to approach this one. Perhaps I am missing something. Here is the problem:

Use continuity to evaluate the limit. Explain your answer (especially why you can use continuity).
$$
\lim_{x \to -2} \frac{x + 2}{\sin\big(\frac{\pi x}{2}\big)}.
$$


We see that this function is not continuous (has a hole) at $-2$ and, for that matter, at all values of $x$ that make $\sin\big(\frac{\pi x}{2}\big) = 0$. We will get $\frac{0}{0}$ if we do all the steps to get down to the point at which we can just plug in $-2$. We cannot use L'Hospital's rule. The best direction I see is something like this:
\begin{align*}
\lim_{x \to -2} \frac{x + 2}{\sin\big(\frac{\pi x}{2}\big)} &= \lim_{x \to -2} \frac{x}{\sin\big(\frac{\pi x}{2}\big)} + \lim_{x \to -2} \frac{2}{\sin\big(\frac{\pi x}{2}\big)}\\
&= \frac{2}{\pi}\lim_{x \to -2} \frac{\frac{\pi x}{2}}{\sin\big(\frac{\pi x}{2}\big)} + 2\lim_{x \to -2} \frac{1}{\sin\big(\frac{\pi x}{2}\big)}
\end{align*}
From there, we might be able to do some $\frac{\sin(x)}{x}$-type thing with the left term, but I am not quite sure what to do with the right term.

 A: I think the right term explodes, and so does the left (because $x\to -2$, the denominator goes to $0$ but the numerator does not). The decomposition performed is therefore incorrect. However,

The key idea is that $-\sin(\pi+x)= \sin x$ for any $x$. This is fairly obvious to prove from the addition formula.

In particular, we have :
$$
\sin\left(\frac{\pi x}{2}\right) = -\sin\left(\frac{\pi x}{2} + \pi\right) = - \sin\left(\frac{\pi (x+2)}{2}\right)
$$
Now, the advantage we have is that we can make use of the $\sin x \over x$ rule, after a change of variable. Indeed, we have :
$$
\frac{x+2}{\sin(\frac{\pi x}{2})} = -\frac{x+2}{\sin\left(\frac{\pi(x+2)}2\right)} = -\frac{2}{\pi} \frac{\frac{\pi(x+2)}{2}}{\sin \left(\frac{\pi(x+2)}{2}\right)}
$$
Let $y = \frac{\pi(x+2)}{2}$. Then, as $x \to -2$ we have $y \to 0$. In particular, $$
\lim_{x \to -2} \frac{x+2}{\sin(\frac{\pi x}{2})} = \lim_{y \to 0} -\frac 2{\pi} \frac{y}{\sin(y)} = \boxed{-\frac{2}{\pi}}
$$
and this can easily be verified from Wolfram Alpha as well.
Note : Thanks to @lalala below. They mention that $\lim_{y\to 0}\frac{y}{\sin y} = 1$ can be proven through L'Hopital. However, we can do better, and that's why this is a standard result.
I would suggest looking at this page for a geometric proof of the inequality $1 < \frac{y}{\sin y} < \frac 1{\cos y}$ for all $y \in (\frac{-\pi }{2} , \frac \pi 2)$(For $y$ negative note that $\sin y = -\sin (-y)$ has the same sign as $y$ so they cancel out). Then the squeeze theorem applies as $y \to 0$ to conclude.
A: We have,
$$\begin{align}\lim_{x \to -2} \frac{x + 2}{\sin\left(\frac{\pi x}{2}\right)}
&=2\lim_{u \to -\pi} \frac{\frac{\pi +u}{\pi}}{-\sin (\pi +u)}\\
&=-2\lim_{u \to -\pi} \frac{\frac{1}{\pi}}{\frac{\sin (\pi +u)}{\pi +u}}\\
&=-\frac {2}{\pi}.\end{align}$$
A: In theses cases a change of coordinates can help, notably by $y=x+2\to 0$ using that $\sin (-\pi + \theta)=-\sin(\theta)$ and standard limits we obtain
$$\lim_{x \to -2} \frac{x + 2}{\sin\left(\frac{\pi x}{2}\right)}=\lim_{y \to 0} \frac{y}{\sin\left(-\pi +\frac{\pi y}{2}\right)}=\lim_{y \to 0} \left(-\frac 2 \pi\right)\cdot \frac{\frac \pi 2y}{\sin\left( \frac{\pi y}{2}\right)}=-\frac 2 \pi$$
