Question

Evaluate $\underset{x\to 4}{lim}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$

• A. $\frac{1}{3}$
• B. $-\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $-\frac{2}{3}$

Answer 1

The correct option is B $-\frac{1}{3}$

$\underset{x\to 4}{lt}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$

First multiply the top and bottom by the conjugate of the numerator and the conjugate of the denominator.

$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\left(\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right)\left(\frac{3+\sqrt{5+x}}{3+\sqrt{5+x}}\right)\left(\frac{1+\sqrt{5-x}}{1+\sqrt{5-x}}\right)$

$=\frac{(9-5-x)(1+\sqrt{5-x})}{(1-5+x)(3+\sqrt{5+x})}$

$=\frac{(4-x)(1+\sqrt{5-x})}{-(4-x)(3+\sqrt{5+x})}$

$=-\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}$

Applying limits

$=-\frac{1+\sqrt{5-4}}{3+\sqrt{5+4}}$

$=\frac{1+1}{3+3}$

$=-\frac{2}{6}$

$=-\frac{1}{3}$.