Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 19|

If f\left(x\right)=\sqrt{x}, find

\lim\limits_{x\to 4}\left(\frac{f\left(x\right)-f\left(4\right)}{x-4}\right)

SOLUTION:

\lim\limits_{x\to 4}\left(\frac{f\left(x\right)-f\left(4\right)}{x-4}\right)=\lim\limits_{x\to 4}\left(\frac{\sqrt{x}-\sqrt{4}}{x-4}\right)

Direct substitution of x=4 gives the indeterminate form \frac{0}{0}. Therefore, we proceed by rationalizing the numerator

=\lim\limits_{x\to 4}\left(\frac{\sqrt{x}-2}{x-4}\right)\cdot \frac{\sqrt{x}+2}{\sqrt{x}+2}

=\lim\limits_{x\to 4}\left(\frac{x-4}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right)

=\lim\limits_{x\to 4}\left(\frac{1}{\sqrt{x}+2}\right)

=\left(\frac{1}{\sqrt{4}+2}\right)

=\frac{1}{4}

 

 

 

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