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

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?

• This is a great conceptional problem in elementary calculus +1 Apr 21 '16 at 0:23

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 $\displaystyle\lim_{x\to 2}f(x)=5$.
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.
• Clever! $+1\,\,\,$ Apr 21 '16 at 0:36