Understanding The Squeeze Theorem I'm having trouble understanding the squeeze theorem, or why, given $f(x)\leq g(x)\leq h(x)$, if the limit of $f(x)$ is $L$ and the limit of $h(x)$ is $L$, the limit of $g(x)$ will also be $L$. 
I concede that if $f(x)=L$ and $h(x)=L$, then $g(x)$ must also have to be $L$. But having a limit doesn't necessarily mean that $f(x)$ and $h(x)$ will be equal to $L$, it just means it approaches $L$. So then $g(x)$ doesn't have to be $L$ then! So why is the squeeze theorem true? 
Can someone please try to explain this to me simply? Thanks. 
 A: The idea is that since $f(x)$ and $h(x)$ both tend to $L$ we can make the differences $f(x)-L$ and $h(x)-L$ as small as we want, then for any $\epsilon >0$
$$f(x)\le g(x)\le h(x) \iff f(x)-L\le g(x)-L\le h(x)-L\iff|g(x)-L|<\epsilon$$
and therefore according to the definition of limit $g(x) \to L$.
A: It's true that $g(x)$ doesn't have to be $l$, but it is enough if it just approaches $l$. Since $f$ and $h$ are both approaching $l$ (as $x$ tends to infinity), and $g$ is "stuck between them", it has no option but to go to $l$ as well.
It's getting squeezed, hence the name!
A: As you've said it doesn't mean that $f$ and $h$ will ever be equal to $L$, however if the limit is $L$ then it means that both $f$ and $h$ attain values arbitrarily close to $L$. Indeed the definition itself gives tells us this, as $\lim_{x \to a} f(x) = L$ means that $\forall \epsilon > 0 \;\exists \delta > 0$ s.t.
$$0<|x-a|<\delta \implies |f(x)-L|< \epsilon$$
Thus intuitively we get that $g(x)$ on a certain neighbourhood around $a$ can be bounded by an arbitrarily small neighbourhood of $L$, which in other words says that $\lim_{x \to a} g(x) = L$
A: In brief, "$h$ tends to $L$" means that you can get values of $h$ as close as you want to $L$.
Then if $f$ and $g$ can both get as close as you want to $L$, $g$ will be at least as close. So $g$ tends to $L$.
