My question arises from when I'm trying to prove that if $\lim_{x\rightarrow a}f(x)=m$ ($m≠0$) then $\lim_{x\rightarrow a}\frac{1}{f(x)}=\frac{1}{m}$ using epsilon delta.
Since $\lim_{x\rightarrow a}f(x)=m$ we have $0<|x-a|<\delta\implies |f(x)-m|<\epsilon_1$.
Now I have to show that $0<|x-a|<\delta\implies \left|\frac{1}{f(x)}-\frac{1}{m}\right|<\epsilon_2$.
$\left|\frac{1}{f(x)}-\frac{1}{m}\right| = \left|\frac{m-f(x)}{mf(x)}\right| < \frac{\epsilon_1}{|m||f(x)|}$.
What I want to say now is that $f(x)≥L$ in the neighbourhood of $x=a$ and set $\epsilon_2= \frac{\epsilon_1}{|m||L|}$ which should complete the proof. However, $f(x)$ is not necessarily a continuous function so doesn't necessarily have a global minimum. But I was wondering, since the limit exists at $a$, wouldn't $f(x)$ at least have a local minimum in the neighbourhood of $x=a$?
As David pointed out, there is no need to complicate the situation if you would like to have a proof of $\lim_{x\to a}\frac1{f(x)}=\frac1{m}$ using epsilon delta.
$$\begin{aligned} &\left|\frac{1}{f(x)}-\frac{1}{m}\right|<\epsilon \\ \impliedby & \left|m-f(x)\right| < \epsilon |m|\, |f(x)|\\ \impliedby & \left|m-f(x)\right| < \epsilon |m|\left|\frac{m}2\right|\text{ and } \left|\frac{m}2\right| < |f(x)|\\ \impliedby & \left|m-f(x)\right| < \epsilon |m|\left|\frac{m}2\right|\text{ and } |m-f(x)| < \left|\frac{m}2\right|\\ \impliedby & \left|m-f(x)\right| < \epsilon_1\\ \end{aligned}$$ where $\epsilon_1=\min\left(\epsilon |m|\left|\frac{m}2\right|, \left|\frac{m}2\right|\right)$. Since $f(x)$ is continuous, there exists $\delta>0$ such that if $|x-a|<\delta$ then $\left|m-f(x)\right| < \epsilon_1$ holds. Our proof is complete.
On the other hand, it might not be easy to determine whether a function can attain a maximum or minimum in an interval in the most general case. For example, there is a function $p(x):\Bbb R\to[-1,1]$ such that $\lim_{x\to0}p(0)=0$ and $p(x)$ does not attain maximum or minimum on any neighbourhood interval of $0$. (We can even make $p(x)$ does not attain maximum nor minimum on any interval except the one-point intervals.) That function is somewhat pathological.
Well, if only continuous function $f(x)$ are considered, then $f(x)$ has a maximum and minimum on every closed interval. If a closed interval is close enough to a given point $x=a$, the maximum and minimum of $f(x)$ on that interval will be close enough to $f(a)$.