# How do you find the limit lim (3-sqrt3^x)/(9-3^x) as x->2?

Jan 7, 2017

$\frac{1}{6}$

#### Explanation:

You can factorize the denominator by the rule:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Then

$9 - {3}^{x} = \left(3 - {\sqrt{3}}^{x}\right) \left(3 + {\sqrt{3}}^{x}\right)$

Let's substitute and simplify:

${\lim}_{x \to 2} \frac{3 - \sqrt{{3}^{x}}}{9 - {3}^{x}} = {\lim}_{x \to 2} \frac{\cancel{3 - \sqrt{{3}^{x}}}}{\cancel{\left(3 - {\sqrt{3}}^{x}\right)} \left(3 + {\sqrt{3}}^{x}\right)} = \frac{1}{3 + {\cancel{\sqrt{3}}}^{\cancel{2}}} = \frac{1}{6}$