Question

The value of $\underset{n\to \infty }{lim}{\left(\frac{a-1+\sqrt[n]{nb}}{a}\right)}^n,(a>0,b>0)$ is equal to

• A. $\sqrt[a]{ab}$
• B. $\sqrt[b]{ba}$
• C. $\sqrt{b}$
• D. $\sqrt{a}$

Answer 1

The correct option is A $\sqrt[a]{ab}$

Here, $\underset{n\to \infty }{lim}{\left(\frac{a-1+\sqrt[n]{nb}}{a}\right)}^n$
$=\underset{n\to \infty }{lim}{(1+\frac{\sqrt[n]{nb}-1}{a})}^n,1^{\infty }$ form
$=e^{\underset{n\to \infty }{lim}\left(\frac{\sqrt[n]{nb}-1}{a}\right)\cdot n}$
$=e^{\underset{n\to \infty }{lim}\frac{1}{a}\cdot \frac{b^{1/n}-1}{1/n}}$
$[\because \underset{n\to \infty }{lim}\frac{b^{1/n}-1}{1/n}=\ln b]$
$=e^{\frac{\ln b}{a}}$
$=e^{\ln b^{1/a}}=b^{1/a}=\sqrt[a]{ab}$