Question

The value of $\underset{x\to \infty }{lim}\left[\frac{(8^{\left(x^n/e^x\right)}-{27}^{\left(x^n/e^x\right)})e^x}{(4^{\left(x^n/e^x\right)}+6^{\left(x^n/e^x\right)}+9^{\left(x^n/e^x\right)})x^n}\right],$ where $n\in N$ is

• A. $\ln \frac{2}{3}$
• B. $\ln \frac{3}{2}$
• C. $\ln \frac{1}{3}$
• D. $\ln 3$

Answer 1

The correct option is A $\ln \frac{2}{3}$

$L=\underset{x\to \infty }{lim}\left[\frac{(8^{\left(x^n/e^x\right)}-{27}^{\left(x^n/e^x\right)})e^x}{(4^{\left(x^n/e^x\right)}+6^{\left(x^n/e^x\right)}+9^{\left(x^n/e^x\right)})x^n}\right]$

Let $\frac{x^n}{e^x}=k$

$\underset{x\to \infty }{lim}\frac{x^n}{e^x}$
Using L'Hospital's Rule
$\underset{x\to \infty }{lim}\frac{x^n}{e^x}=\underset{x\to \infty }{lim}\frac{n!}{e^x}=0$
$\Rightarrow k\to 0$ as $x\to \infty$

$\Rightarrow L=\underset{k\to 0}{lim}\left[\frac{(8^k-{27}^k)}{(4^k+6^k+9^k)k}\right]$

$\Rightarrow L=\underset{k\to 0}{lim}\frac{(2^k-3^k)}{k}$
Using L'Hospital's Rule
$\Rightarrow L=\ln 2-\ln 3=\ln \frac{2}{3}$