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

$\lim _{x \rightarrow 0}\left(\frac{\sinh 2 x}{2 x}\right)^{\frac{1}{x^2}} = $

$\lim _{n \rightarrow \infty} \frac{3 \cdot 2^{n+1}-4 \cdot 5^{n+1}}{5 \cdot 2^{n}+7 \cdot 5^{n}}$ is equal to

If $f(x) = \frac{\sin(e^{x-2} - 1)}{\log(x-1)},$ then $\lim_{x \to 2} f(x)$ is given by

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $

If $\beta = \lim_{x \rightarrow 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$,then the value of $6 \beta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module