The Squeeze Theorem

Near $a$, each value $f(x)$ is caught between $g(x)$ and $h(x)$, and both of those approach $L$. So $f(x)$ has nowhere else to go. The theorem is the tool for a function whose limit no algebra reaches, usually one that oscillates.

Substituting $0$ gives $\frac{0}{0}$, and no algebra cancels it. The proof is a squeeze, read off the unit circle centered at $O$. Take an angle $x$ with $0 < x < \frac{\pi}{2}$. Let $A = (1, 0)$, let $P$ be the point at angle $x$ on the circle, so $P = (\cos x, \sin x)$, and let $T$ be where the ray $OP$ meets the vertical line through $A$.

The triangle $OAP$ has base $1$ and height $\sin x$. The sector has radius $1$ and angle $x$, so its area is $\frac{x}{2}$. That's where radians are needed: in degrees the area would be $\frac{\pi x}{360}$. The triangle $OAT$ has a right angle at $A$ and base $OA = 1$, so its height is $AT = \tan x$. Each region lies inside the next, so

Multiply by $\frac{2}{\sin x}$, which is positive, so the inequalities keep their direction: $1 \le \frac{x}{\sin x} \le \frac{1}{\cos x}$. All three are positive, so taking reciprocals reverses them:

For $-\frac{\pi}{2} < x < 0$, replacing $x$ with $-x$ changes neither $\cos x$ nor $\frac{\sin x}{x}$, so the same inequalities hold there. As $x$ approaches $0$, $\cos x$ approaches $1$, and the squeeze theorem gives the limit $1$.