Question

$\underset{x\to \infty }{lim}{(e^{11x}-7x)}^{\frac{1}{3x}}=e^{\lambda }$, (where [.] denotes greatest integer function), then $\left[\lambda \right]$ is equal to

Answer 1

Let $y=\underset{x\to \infty }{lim}{(e^{11x}-7x)}^{\frac{1}{3x}}$

$\ln y=\underset{x\to \infty }{lim}\frac{\ln (e^{11x}-7x)}{3x}$

$\Rightarrow \ln y=\underset{x\to \infty }{lim}\frac{(11e^{11x}-7)}{(e^{11x}-7x)3}=\underset{x\to \infty }{lim}\frac{11-\frac{7}{e^{11x}}}{3-\frac{21x}{e^{11x}}}$

$\ln y=\frac{11}{3}$
$y=e^{\frac{11}{3}}$

$\therefore \left[\lambda \right]=\left[\frac{11}{3}\right]=3$