Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 7

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PROBLEM:

Evaluate lim⁡x→1 x−1x+3−2 \displaystyle \lim\limits_{x\to 1}\:\frac{x-1}{\sqrt{x+3}-2}.

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 SOLUTION:

A straight substitution of  x=1x=1 leads to the indeterminate form 00 \frac{0}{0} which is meaningless.

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

lim⁡x→1 x−1x+3−2=lim⁡x→ 1 x−1x+3−2⋅x+3+2x+3+2=lim⁡x→1(x−1)(x+3+2)(x+3)−22=lim⁡x→1(x−1)(x+3+2)x−1=lim⁡x→1x+3+2=1+3+2=4+2=2+2=4  (Answer)\begin{align*} \\ \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 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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