Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 11

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Evaluate lim⁡x→3(x−3x−2−4−x) \displaystyle \lim\limits_{x\to 3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)

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Solution:

A straight substitution of x=3x=3 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→ 3(x−3x−2−4−x)=lim⁡x→  3(x−3x−2−4−x)⋅x−2+4−xx−2+4−x=lim⁡x→3[(x−3)(x−2+4−x)(x−2−4−x)(x−2+4−x)]=lim⁡x→3[(x−3)(x−2+4−x)2x−6]=lim⁡x→3[(x−3)(x−2+−x+4)2(x−3)]=lim⁡x→3[x−2+4−x2]=3−2+4−32=1  (Answer)\begin{align*} \begin{align*} \lim\limits_{x\to \:3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right) & =\lim\limits_{x\to \:\:3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)\cdot \frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}} \\ & =\lim\limits_{x\to 3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{\left(\sqrt{x-2}-\sqrt{4-x}\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}\right] \\ & =\lim\limits_{x\to3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{2x-6}\right]\\ & =\lim\limits_{x\to3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{-x+4}\right)}{2\left(x-3\right)}\right] \\ & =\lim\limits_{x\to 3}\left[\frac{\sqrt{x-2}+\sqrt{4-x}}{2}\right]\\ & =\frac{\sqrt{3-2}+\sqrt{4-3}}{2}\\ & =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*} \end{align*}
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