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

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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 $g$.

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  • $\begingroup$ 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 $\delta=\epsilon \cdot (M-f(x))\cdot (M-f(a))$? $\endgroup$ – W. G. Dec 3 at 1:37
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    $\begingroup$ Nevermind this answers my question. Thank you. $\endgroup$ – W. G. Dec 3 at 1:49

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