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

The value of $\lim _{n \rightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$ is

$\lim _{n}$ ${\rightarrow \infty} \sqrt{2} \left[ \frac{(2+\sqrt{2})^n + (2-\sqrt{2})^n}{(2+\sqrt{2})^n - (2-\sqrt{2})^n} \right] =$

$\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sqrt{15+\cos 2x}-4} = $

Evaluate the limit: $\lim _{n \rightarrow \infty} \frac{A+e^{n x}}{x+A e^{n x}}$

$\lim _{x \rightarrow \infty} \frac{3 x+4 \cos ^2 x}{\sqrt{x^2-5 \sin ^2 x}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module