Question

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