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How to find $\lim\limits_{x\to 2}f(x)$ if $\lim\limits_{x\to 2}\frac{f(x)-5}{x-2}=100$?

I suppose that we do not use L'Hopital rule here. Then $\lim\limits_{x\to 2}f(x)-5=0$, then $\lim\limits_{x\to 2}f(x)=5.$

I highly doubt the answer, could someone help me understand?

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    $\begingroup$ This is a great conceptional problem in elementary calculus +1 $\endgroup$ – imranfat Apr 21 '16 at 0:23
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Note that the numerator must be $0$ in order to obtain the indeterminate form, otherwise the limit will just be $\pm \infty$, thus we must have $\displaystyle\lim_{x\to 2}f(x)=5$.

So yes, your answer is correct.

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First write $$\lim\limits_{x \to 2} f(x)=\lim\limits_{x \to 2}\left[\frac{f(x)-5}{x-2}(x-2)+5\right]$$ and since three limits exist $$=\lim\limits_{x \to 2}\frac{f(x)-5}{x-2}\lim\limits_{x \to 2}(x-2)+\lim\limits_{x \to 2}5=100\cdot0+5=5.$$

Proof of the statement that EdwardJiang used: If $$\lim\limits_{x \to a} \frac{f(x)}{g(x)}=b \text{ and } \lim\limits_{x \to a} g(x)=0$$ then $$\lim\limits_{x \to a} f(x)=\lim\limits_{x \to a} \frac{f(x)}{g(x)}g(x)=\text{(limits exist)}\lim\limits_{x \to a} \frac{f(x)}{g(x)}\lim\limits_{x \to a} g(x)=b\cdot 0=0.$$ (Maybe I am wrong but) I disagree about using this information without proof in this question. This question is actually about showing this as an example.

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    $\begingroup$ Clever! $+1\,\,\,$ $\endgroup$ – Cameron Williams Apr 21 '16 at 0:36
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    $\begingroup$ This is exactly the kind of answer which shows the power of rules of algebra of limits. Most intuitive statement about limits (if ratio tends to a limit and denominator tends to zero then numerator must also tend to zero) are an immediate consequence of algebra of limits and students should make good use of algebra of limits. +1 $\endgroup$ – Paramanand Singh Apr 21 '16 at 10:26

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