## Calculus: Early Transcendentals (2nd Edition)

$f(x)=x^2-1$
We have to find function $f(x)$ so that $\lim\limits_{x \to 1}\left(\dfrac{f(x)}{x-1}\right)=2$. In order for the limit to exist $f(x)$ should have a factor $(x-1)$. We can choose any function $g$ so that we have: $f(x)=(x-1)g(x)$ $g(1)=2$ For example: $g(x)=(x+1)\Rightarrow f(x)=(x-1)(x+1)=x^2-1$ $\lim\limits_{x \to 1}\left(\dfrac{x^2-1)}{x-1}\right)=\lim\limits_{x \to 1}\left(\dfrac{(x-1)(x+1)}{x-1}\right)=\lim\limits_{x \to 1} (x+1)=1+1=2$