Question

The value of $2\left(\displaystyle\lim _{x \rightarrow 0^{-}} \frac{4^{\frac{3}{x}}+15\left(2^{\frac{1}{x}}\right)}{\left(1+\frac{6}{x}\right)+6\left(2^{\frac{1}{x}}\right)}\right)$ is equal to

Answer 1

$\displaystyle\lim _{x \rightarrow 0^{-}} \frac{2^{\frac{6}{x}}+15 \cdot 2^{\frac{1}{x}}}{2^{1+\frac{6}{x}}+6 \cdot 2^{\frac{1}{x}}}$
$=\displaystyle\lim _{x \rightarrow 0^{-}} \frac{2^{\frac{1}{x}}\left\{2^{\frac{5}{x}}+15\right\}}{\left.2^{\frac{1}{x}}\left\{2^{\frac{5}{x}}\right)+6\right\}}$
$=\displaystyle\lim _{x \rightarrow 0^{-}} \frac{2^{\frac{5}{x}}+15}{2\left(2^{\frac{5}{x}}\right)+6}$
$=\frac{0+15}{2(0)+6}=\frac{15}{6}=\frac{5}{2}$
$\therefore 2\left(\displaystyle\lim _{x \rightarrow 0^{-}} \frac{4^{\frac{3}{x}}+15\left(2^{\frac{1}{x}}\right)}{2^{\left(1+\frac{6}{x}\right)}+6\left(2^{\frac{1}{x}}\right)}\right)=5$