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Evaluate \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=3 leads to the indeterminate form \frac{0}{0} which is meaningless.
Therefore, to evaluate the limit of the given function, we proceed as follows.
\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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