Explanation Of The Sandwich Theorem

Question

I know what the sandwich is - it states if we take 2 functions, one which gives consistently greater values than f(x), and one which gives consistently lower values than f(x), if the converge on the same point, then f(x) will converge there too.

But can someone explain to me why this is true? I know there are epsilon delta proofs on wikipedia and such, but that is far too rigorous for me. Is it possible, for someone to explain to me simply and intuitively why the sandwich theorem is true?

Thanks.

Answer 1

$g(x) \to k$ means $|g(x) - k|$ "is small" for "big $x$".

And $h(x)\to k$ mean $|h(x)-k|$ "is small" for "big $x$".

But $g(x) < f(x) < h(x)$. Where does $k$ fit in.

If $k \le f(x)$ then $|f(x) - k| \le |h(x) - k|$ so $|f(x) - k|$ is also small.

If $k \ge f(x)$ then $|f(x) - k| \le |g(x) - k|$ so $|f(x) - k|$ is also small.

So no matter what, if $|g(x)-k|$ and $|h(x)-k|$ are small and $g(x) < f(x) < h(x)$ for all $x$ then $|f(x)- k|$ is going to be at least as small as one of them.

So $f(x) \to k$.

Now that's not rigorous because we haven't really defined what $f(x)\to k$ means and what $|g(x) -k|$ is "small" means.

But no matter how we eventually do define it, it will always follow from the definition that $f(x)$ and $k$ are at least as close as either $g(x)$ or $h(x)$ is to $k$.

Answer 2

Intuitively, a function converges to a limit if it gets closer and closer to it. Now it you have two functions converging to the same limit and a third one bracketed between the first two, it has no escape, it must converge to that limit as well.

---

In terms of intervals, for any $\delta$ there is an $\epsilon$* such that

$$L-\epsilon<f<L+\epsilon$$ and $$L-\epsilon<g<L+\epsilon.$$

With $f<h<g$ this obviously implies

$$L-\epsilon<f<h<g<L+\epsilon$$ thus $$L-\epsilon<h<L+\epsilon.$$

*Technically, the $\epsilon$ might be different for $f$ and $g$, but it suffices to take the largest of the two.