Question

The value of $\lim _{x \rightarrow 2} \frac{3^{x / 2}-3}{3^x-9}$ is

(a) 0

(b) 1/3

(c) $1 / 6$

(d) $\log 3$

Answer 1

$\begin{aligned}\lim _{x \rightarrow 2}\left(\frac{3^{x / 2}-3}{3^x-9}\right) & =\lim _{x \rightarrow 2}\left(\frac{3^{x / 2}-3}{\left(3^{x / 2}\right)^2-3^2}\right) \\& =\lim _{x \rightarrow 2} \frac{1}{\left(3^{x / 2}+3\right)} \\& =\frac{1}{3^1+3}=\frac{1}{6}\end{aligned}$

So, The correct option of this question will be (C).