# Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 22|

### $\lim\limits_{x\to 0}\left(\frac{f\left(x+2\right)-f\left(2\right)}{x}\right)$$\lim\limits_{x\to 0}\left(\frac{f\left(x+2\right)-f\left(2\right)}{x}\right)$

SOLUTION:

$\lim\limits_{x\to 0}\left(\frac{f\left(x+2\right)-f\left(2\right)}{x}\right)=\lim\limits_{x\to 0}\left(\frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(2^2-2\cdot 2+3\right)}{x}\right)$

$=\lim\limits_{x\to 0}\left(\frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(3\right)}{x}\right)$

$=\lim\limits_{x\to 0}\left(\frac{\left(\left(x+2\right)^2-2\left(x+2\right)\right)}{x}\right)$

Direct substitution of $x=0$ gives the indeterminate form $\frac{0}{0}$. Therefore, we proceed by factoring the numerator

$=\lim\limits_{x\to 0}\frac{\left(x+2\right)\left(x+2-2\right)}{x}$

$=\lim\limits_{x\to 0}\frac{\left(x+2\right)\left(x\right)}{x}$

$=\lim\limits_{x\to 0}\left(x+2\right)$

$=0+2$

$=2$