Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 6|

Evaluate

\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}

 SOLUTION:

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{\sqrt{x+16}-4}{x}=\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}\cdot \frac{\sqrt{x+16}+4}{\sqrt{x+16}+4}

=\lim\limits_{x\to 0}\:\frac{\left(x+16\right)-4^2}{x\left(\sqrt{x+16}+4\right)}

=\lim\limits_{x\to 0}\:\frac{x+16-16}{x\left(\sqrt{x+16}+4\right)}

=\lim\limits_{x\to 0}\:\frac{x}{x\left(\sqrt{x+16}+4\right)}

=\lim\limits_{x\to 0}\:\frac{1}{\sqrt{x+16}+4}

=\:\frac{1}{\sqrt{0+16}+4}

=\:\frac{1}{4+4}

=\:\frac{1}{8}

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