Suppose for all $x$ we know $g(x)\le f(x)\le h(x)$ and $\lim_{x\to c} g(x)=L=\lim_{x\to c} h(x)$. Does the following argument work to conclude that $\lim_{x\to c} f(x)=L$?

Let $\epsilon\gt 0$ be given. Then we can find a $\delta_1$ such that if $|x-c|\lt\delta_1$, then $|g(x)-L|\lt\epsilon$ and a $\delta_2$ such that if $|x-c|\lt\delta_2$ then $|h(x)-L|\lt\epsilon$.

Let $\delta = min\{\delta_1,\delta_2\}$. Then for all x such that $|x-c|\lt\delta$, it follows that $|g(x)-L|\lt\epsilon$ and $|h(x)-L|\lt\epsilon$.

This means that $L-\epsilon\lt g(x)\le h(x)\lt L+\epsilon$. But since $g(x)\le f(x)\le h(x)$, $L-\epsilon\lt f(x) \lt L+\epsilon$. Hence $\lim_{x\to c}f(x)=L$.

This proof is slightly different from others that I've seen, but it doesn't seem to be wrong. Is there anything that I'm missing?

  • $\begingroup$ What proof have you seen before? $\endgroup$
    – Mark Viola
    Mar 17, 2015 at 2:46
  • $\begingroup$ You have a typo in the condition. It should be $\lim_{x\to c} g(x) = L = \lim_{x\to c} h(x)$. $\endgroup$ Mar 17, 2015 at 9:42

1 Answer 1


This looks like a good proof. I don't see anything wrong with it.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .