# Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 12|

#### $\lim\limits_{x\to 0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right)$

SOLUTION:

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

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

$\lim\limits_{x\to \:0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right)=\lim\limits_{x\to 0}\left(\frac{1\cdot \:\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)}{x}\right)$

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

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

$=\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right)\cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}$

$=\lim\limits_{x\to 0}\left(\frac{x}{3x\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right)$

$=\lim\limits_{x\to 0}\left(\frac{1}{3\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right)$

$=\frac{1}{3\left(\sqrt{0+9}+3\right)\sqrt{0+9}}$

$=\frac{1}{54}$