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. $\sqrt[a]{aa}$
• C. $\sqrt{b}$
• D. $\sqrt{a}$

Answer 1

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

${\left(\frac{a-1+\sqrt[n]{nb}}{a}\right)}^n=y$

$y=e^{n\log \left(\frac{a-1+\sqrt[n]{nb}}{a}\right)}$

$\underset{n\to \infty }{lim}y=e^{\underset{n\to \infty }{lim}n\log \left(\frac{a-1+b^{\frac{1}{n}}}{a}\right)}$

Replace $n$ with $\frac{1}{h}$, where as $n\to \infty ,h\to 0$,

$\therefore \underset{n\to \infty }{lim}y=e^{\underset{h\to 0}{lim}\frac{\log \left(\frac{a-1+b^h}{a}\right)}{h}}$

Using L'Hospital's Rule,
$\therefore \underset{n\to \infty }{lim}y=e^{\underset{h\to 0}{lim}\frac{b^h\log b}{a-1+b^h}}$

$y=e^{\frac{logb}{a}}$

$y=\sqrt[a]{ab}$