Question

If $\underset{x\to \infty }{lim}{(e^{2x}+e^x+x)}^{\frac{1}{x}}=e^a$, then the value of $a$ is

Answer 1

$L=e^{\underset{x\to \infty }{lim}\frac{\ln (e^{2x}+e^x+x)}{x}}$
Using L-Hospital's rule
$\Rightarrow e^{\underset{x\to \infty }{lim}\frac{2e^{2x}+e^x+1}{e^{2x}+e^x+x}}=e^a$
Now, dividing by $e^{2x}$ and since we know $x\to \infty \Rightarrow e^{-2x}\to 0$
So, we have $e^a=e^{\left(\frac{2+0+0}{1+0+0}\right)}$
$\therefore e^2=e^a\Rightarrow a=2$