Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 12

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PROBLEM:

Evaluate lim⁡x→0(1x(13−1x+9))\displaystyle \lim\limits_{x\to 0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right)

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SOLUTION: 

A straight substitution of x=0x=0 leads to the indeterminate form 0⋅00\cdot 0 which is meaningless.

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

lim⁡x→ 0(1x(13−1x+9))=lim⁡x→0(1⋅ (13−1x+9)x)=lim⁡x→0(x+9−33x+9x)=lim⁡x→0(x+9−33xx+9)=lim⁡x→0(x+9−33xx+9)⋅x+9+3x+9+3=lim⁡x→0(x3x(x+9+3)x+9)=lim⁡x→0(13(x+9+3)x+9)=13(0+9+3)0+9=154  (Answer)\begin{align*} \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} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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