Question

Find the value of $\underset{x\to 2}{lim}\frac{3^{\frac{x}{2}}-3}{3^x-9}$.

Answer 1

We can write

$3^x-9=(3^{\frac{x}{2}}+3)(3^{\frac{x}{2}}-3)$ [since $a^2-b^2=(a+b)(a-b)$].

$\therefore \underset{x\to 2}{lim}\frac{(3^{\frac{x}{2}}-3)}{(3^x-9)}=\underset{x\to 2}{lim}\frac{(3^{\frac{x}{2}}-3)}{(3^{\frac{x}{2}}-3)(3^{\frac{x}{2}}+3)}=\underset{x\to 2}{lim}\frac{1}{(3^{\frac{x}{2}}+3)}$.

Now substituting limits we get $\frac{1}{3+3}=1/6$.