Differential and Integral Calculus| Feliciano and Uy| Limit of a Function in Indeterminate Form|Exercise 1.3| Problem 5|

$\lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-9}{2x}$

SOLUTION:

A straight substitution of $x=0$ leads to the indeterminate form  $\frac{0}{0}$  which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

$\lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-9}{2x}=\:\lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-\left(3\right)^2}{2x}$

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

$=\lim\limits_{x\to 0}\:\frac{x\left(x+6\right)}{2x}$

$=\lim\limits_{x\to 0}\:\frac{x+6}{2}$

$=\frac{0+6}{2}$

$=\frac{6}{2}$

$=3$