Question

$\underset{x \rightarrow \infty}{\text{Lim}} \left(e^{11 x}-7 x\right)^{\frac{1}{3 x}}$ is equal to

• A. $\frac{11}{3}$
• B. $\frac{3}{11}$
• C. $e ^{\frac{3}{11}}$
• D. $e ^{\frac{11}{3}}$

Answer 1

Answer: (D)

Let $L=\underset{x \rightarrow \infty}{\text{Lim}} \left(e^{11 x}-7 x\right)^{1 / 3 x} \left(\infty^{\circ}\right.$ form $)$
$\left(\operatorname{As} \underset{x \rightarrow \infty}{\text{Lim}}\left(e^{11 x}-7 x\right)=\underset{x \rightarrow \infty}{\text{Lim}}\left(1-\frac{7 x}{e^{11 x}}\right) e^{11 x}=(1-0) \times \infty \rightarrow \infty \text { and } \underset{x \rightarrow \infty}{\text{Lim}} \frac{1}{3 x} \rightarrow 0\right) $
$\ln L=\underset{x \rightarrow \infty}{\text{Lim}} \frac{\ln \left(e^{11 x}-7 x\right)}{3 x}$
(Using L-Hospital rule)
$\ln L=\underset{x \rightarrow \infty}{\text{Lim}} \frac{\left(11 e^{11 x}-7\right)}{3\left(e^{11 x}-7 x\right)}=\underset{x \rightarrow \infty}{\text{Lim}} \frac{\left(11-7 e^{-11 x}\right)}{3\left(1-7 x e^{-11 x}\right)} . \text { Hence } L=e^{\frac{11}{3}}$