If a 0, b 0,then lim _{n rightarrow infty | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $L = \lim _{n \rightarrow \infty}\left(\frac{a + b^{1 / n} - 1}{a}\right)^n$. Taking the natural logarithm on both sides: $\ln L = \lim _{n \r

Vedclass · Accepted Answer

$b^{1 / a}$

Vedclass · Answer

(C) Let $L = \lim _{n \rightarrow \infty}\left(\frac{a + b^{1 / n} - 1}{a}\right)^n$. Taking the natural logarithm on both sides: $\ln L = \lim _{n \rightarrow \infty} n \ln \left(1 + \frac{b^{1 / n} - 1}{a}\right)$. Using the limit formula $\lim _{x \rightarrow 0} \frac{\ln(1 + x)}{x} = 1$,let $x = \frac{b^{1 / n} - 1}{a}$. As $n \rightarrow \infty$,$x \rightarrow 0$. $\ln L = \lim _{n \rightarrow \infty} n \cdot \left(\frac{b^{1 / n} - 1}{a}\right) \cdot \frac{\ln(1 + x)}{x} = \lim _{n \rightarrow \infty} \frac{n}{a} (b^{1 / n} - 1)$. Let $t = 1/n$. As $n \rightarrow \infty$,$t \rightarrow 0$. $\ln L = \frac{1}{a} \lim _{t \rightarrow 0} \frac{b^t - 1}{t}$. Since $\lim _{t \rightarrow 0} \frac{b^t - 1}{t} = \ln b$,we have: $\ln L = \frac{1}{a} \ln b = \ln(b^{1 / a})$. Therefore,$L = b^{1 / a}$.

If $a > 0, b > 0$,then $\lim _{n \rightarrow \infty}\left(\frac{a + b^{1 / n} - 1}{a}\right)^n =$

Similar Questions

If $f(x) = \begin{cases} x^2 - 3, & 2 < x < 3 \\ 2x + 5, & 3 < x < 4 \end{cases}$,the equation whose roots are $\lim_{x \to 3^-} f(x)$ and $\lim_{x \to 3^+} f(x)$ is

$\mathop {\lim }\limits_{x \to \pi /2} \frac{{1 + \cos 2x}}{{{{(\pi - 2x)}^2}}} = $

The limit of the interior angle of a regular polygon of $n$ sides as $n \rightarrow \infty$ is

If $0 \leq x \leq \pi / 2$,then $\lim _{x \rightarrow a} \frac{|2 \cos x-1|}{2 \cos x-1}$

Find the limit: $\mathop {\lim }\limits_{x \to 3} [x(x+1)]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module