Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 11|

Evaluate

\lim\limits_{x\to 3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)

 

 

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

\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\to 3}\left(\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{2x-6}\right)

=\lim\limits_{x\to 3}\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

 

 

 

 

 

 

 

 

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