Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 7|

Evaluate

\lim\limits_{x\to 1}\:\frac{x-1}{\sqrt{x+3}-2}

 SOLUTION:

A straight substitution of  x=1 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 1}\:\frac{x-1}{\sqrt{x+3}-2}=\lim\limits_{x\to \:1}\:\frac{x-1}{\sqrt{x+3}-2}\cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}

=\lim\limits_{x\to 1}\frac{\left(x-1\right)\left(\sqrt{x+3}+2\right)}{\left(x+3\right)-2^2}

=\lim\limits_{x\to 1}\frac{\left(x-1\right)\left(\sqrt{x+3}+2\right)}{x-1}

=\lim\limits_{x\to 1}\sqrt{x+3}+2

=\sqrt{1+3}+2

=\sqrt{4}+2

=2+2

=4

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