mathop {lim }limits_{x to infty } ,{left( {fr | vedclass

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

Question

(C) We have the limit $\mathop {\lim }\limits_{x 	o \infty } \,{\left( {\frac{{x + a}}{{x + b}}} ight)^{x + b}}$.Rewrite the expression inside the

Vedclass · Accepted Answer

$e^{a - b}$

Vedclass · Answer

(C) We have the limit $\mathop {\lim }\limits_{x 	o \infty } \,{\left( {\frac{{x + a}}{{x + b}}} ight)^{x + b}}$.Rewrite the expression inside the limit as $\mathop {\lim }\limits_{x 	o \infty } \,{\left( {1 + \frac{{a - b}}{{x + b}}} ight)^{x + b}}$.Using the standard limit formula $\mathop {\lim }\limits_{u 	o \infty } (1 + \frac{k}{u})^u = e^k$,where $u = x + b$ and $k = a - b$:$\mathop {\lim }\limits_{x 	o \infty } \,{\left( {1 + \frac{{a - b}}{{x + b}}} ight)^{x + b}} = e^{a - b}$.Thus,the correct option is $C$.

$\mathop {\lim }\limits_{x \to \infty } \,{\left( {\frac{{x + a}}{{x + b}}} \right)^{x + b}} = $

Similar Questions

For $x \in R$,$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x - 3}}{{x + 2}}} \right)^x}$ is equal to

$\mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}} + \frac{3}{{{n^2}}} + \dots + \frac{n}{{{n^2}}}} \right\}$ is

If $\mathop {\lim }\limits_{x \to 0} \frac{{\log (a + x) - \log a}}{x} + k\mathop {\lim }\limits_{x \to e} \frac{{\log x - 1}}{{x - e}} = 1$,then

$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals

The limit of $\left[\frac{1}{x^{2}}+\frac{(2013)^{x}}{e^{x}-1}-\frac{1}{e^{x}-1}\right]$ as $x \rightarrow 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module