Question

Limits that lead to the indeterminate forms $1^{\infty },0^0,{\infty }^0$ can sometimes be solved taking logarithm first and then using L' Hospital's rule

Let $\underset{x\to a}{lim}\left(f\right(x){)}^{g\left(x\right)}$ is in the form of ${\infty }^0$, it can be written as $e^{{lim}_{x\to a}g\left(x\right)\ln f\left(x\right)}=e^L$ where $L=\underset{x\to a}{lim}\frac{\ln f\left(x\right)}{\frac{1}{g\left(x\right)}}$ is $\frac{\infty }{\infty }$ form and can be solved using L' Hopital's rule.

The value of $\underset{x\to \infty }{lim}\left[\right(\ln x{)}^{\frac{1}{2x}}+x^{\frac{1}{x^n}}]\forall n\in N$ is :

• A. $2$
• B. $0$
• C. $e^{\frac{1}{2}}$
• D. $e$

Answer 1

The correct option is A $2$

$\underset{x\to \infty }{lim}\left[\right(\ln x{)}^{\frac{1}{2x}}+x^{\frac{1}{x^n}}]=\underset{x\to \infty }{lim}(\ln x{)}^{\frac{1}{2x}}+\underset{x\to \infty }{lim}{x}^{\frac{1}{x^n}}\to \left(1\right)$

Let,

$m=\underset{x\to \infty }{lim}(\ln x{)}^{\frac{1}{2x}}\&n=x^{\frac{1}{x^n}}$

Taking $\log$ on both sides on both equations,
$\log m=\underset{x\to \infty }{lim}\frac{\log \left(\ln x\right)}{2x}$ and $\log n=\underset{x\to \infty }{lim}\frac{logx}{x^n}$
Clearly, both the limits are of form $\frac{\infty }{\infty }$, and applying L'hopital's rule,
$\log m=\underset{x\to \infty }{lim}\frac{\frac{1}{x}}{2\ln x}$ and $\log n=\underset{x\to \infty }{lim}\frac{\frac{1}{x}}{n{x}^{n-1}}$

$\Rightarrow \log m=0\&\log n=0$
$\Rightarrow m=1\&n=1$

Putting in (1), we get,
$\underset{x\to \infty }{lim}\left[\right(\ln x{)}^{\frac{1}{2x}}+x^{\frac{1}{x^n}}]=\underset{x\to \infty }{lim}(\ln x{)}^{\frac{1}{2x}}+\underset{x\to \infty }{lim}{x}^{\frac{1}{x^n}}=1+1=2$