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

Differential and Integral Calculus by Feliciano and Uy Solution Manual by


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


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}