# Proof Question-Extreme Value Theorem (Continuity of Functions)

I was looking at the following proof of the Extreme Value Theorem.

$$\textbf{Question:}$$ Why does $$f(x)$$ is continuous and $$g(x)=\frac{1}{M-f(x)} \implies g(x)$$ is continuous?

We know $$f(x)$$ is continuous here which means when $$a\in X$$

$$\forall \epsilon > 0$$ $$\exists \delta >0$$ $$\forall x\in X$$, $$|x-a|<\delta\rightarrow |f(x)-f(a)|<\epsilon$$.

My Attempt to Try to Prove $$g(x)$$ is Continuous: Let $$\epsilon >0$$. Working backward to find a $$\delta$$.

We want to show $$|g(x)-g(a)|<\epsilon$$. In other words, $$-\epsilon. Continuing, $$-\epsilon<\frac{1}{M-f(x)}-\frac{1}{M-f(a)}<\epsilon$$. I don't know where to go from here...

Follows immediately from actually simplifying that expression that you have for $$g(x)-g(a)$$. In the numerator, you will get a term that goes to $$0$$ due to the continuity of $$f$$, and the denominator is not infinitesimal (note that they define $$M>f$$). Then, for a given pair $$(\epsilon,\delta)$$ for $$f$$, you will be able to find a corresponding $$(\epsilon,\delta')$$ for $$g$$.
• Doing that I get $\frac{1}{M-f(x)}-\frac{1}{M-f(a)}=\frac{M-f(a)-(M-f(x))}{(M-f(x))\cdot (M-f(a))}=\frac{f(x)-f(a)}{(M-f(x))\cdot (M-f(a))}\implies -\epsilon\cdot (M-f(x))\cdot (M-f(a))<f(x)-f(a)<\epsilon \cdot (M-f(x))\cdot (M-f(a))$. Then, would I take $\delta=\epsilon \cdot (M-f(x))\cdot (M-f(a))$? – W. G. Dec 3 at 1:37