Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 20

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PROBLEM:

If f(x)=x \displaystyle f\left(x\right)=\sqrt{x}, find lim⁡x→0(f(9+x)−f(9)x)\displaystyle \lim\limits_{x\to 0}\left(\frac{f\left(9+x\right)-f\left(9\right)}{x}\right).

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SOLUTION:

lim⁡x→0(f(9+x)−f(9)x)=lim⁡x→0(9+x−9x)\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(9+x\right)-f\left(9\right)}{x}\right)=\lim\limits_{x\to 0}\left(\displaystyle \frac{\sqrt{9+x}-\sqrt{9}}{x}\right)

Direct substitution of x=0x=0 gives the indeterminate form 00\frac{0}{0}. Therefore, we proceed by rationalizing the numerator.

=lim⁡x→0(9+x−3x⋅9+x+39+x+3)=lim⁡x→0(9+x−9x(9+x+3))=lim⁡x→0(xx(9+x+3))=lim⁡x→0(1(9+x+3))=(1(9+0+3))=16  (Answer)\begin{align*} & =\lim\limits_{x\to 0}\left(\displaystyle \frac{\sqrt{9+x}-3}{x}\cdot \displaystyle \frac{\sqrt{9+x}+3}{\sqrt{9+x}+3}\right)\\ \\ & =\lim\limits_{x\to 0}\left(\displaystyle \frac{9+x-9}{x\left(\sqrt{9+x}+3\right)}\right)\\ \\ & =\lim\limits_{x\to 0}\left(\displaystyle \frac{x}{x\left(\sqrt{9+x}+3\right)}\right)\\ \\ & =\lim\limits_{x\to 0}\left(\displaystyle \frac{1}{\left(\sqrt{9+x}+3\right)}\right)\\ \\ & =\left(\displaystyle \frac{1}{\left(\sqrt{9+0}+3\right)}\right)\\ \\ & =\displaystyle \frac{1}{6} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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