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Let the functions $f(x)$ and $g(x)$ are defined and differentiable in the interval $(a,b)$ and everywhere in it $g'(x)\ne 0.$ Let $$\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0.$$ Let also the limit of the fraction of the derivatives $$ \lim_{x\to a}\frac{f'(x)}{g'(x)} $$ exists in a generalised sense. Then $$ \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)} $$

The proof of that theorem starts by saying that we can extend the continuity of the functions $f(x)$ and $g(x)$ in the point $a$ by denoting $f(a)=g(a)=0$ and preserve the same denotations for the extended functions.

I really don't get that. The function is continuous in the interval $(a,b)$ from which I conclude that it can be undefined in the point a. How are we allowed to say that $f(a)=g(a)=0$?

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Suppose we define $F$ on $[a,b)$ by setting $F(x) = f(x), x\in (a,b),$ $F(a)=0.$ Do the same with $G,g.$ Then $F,G$ are continuous on $[a,b).$ These functions are extensions of the orginal functions. You're right, it doesn't really make strict sense to talk of $f(a),g(a),$ but this process is so simple and natural that we usually drop the $F,G$ business and use $f,g$ for the extensions as well.

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  • $\begingroup$ But if we prove the Theorem for the extended functions does it mean the same theorem will be true for the subfunctions? $\endgroup$ – Nikola Oct 1 '17 at 17:45
  • $\begingroup$ Yes, because limits don't care about the actual value of the function at the limiting point. $\endgroup$ – themathandlanguagetutor Oct 1 '17 at 18:07
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    $\begingroup$ Sure, just note $f(x)/g(x) = (F(x)-F(a))/(G(x)-G(a))$ and the various MVT's apply to $F,G$ and furthermore $F'(x)= f'(x), G'(x) = g'(x)$ for $x\in (a,b).$ $\endgroup$ – zhw. Oct 1 '17 at 18:20

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