If A neq 0 and x 0,then lim _{n rightarrow | vedclass

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

Question

(D) We are given the limit: $\lim _{n \rightarrow \infty} \frac{\cos x - e^{nx}}{1 - A e^{nx}}$.Divide the numerator and the denominator by $e^{nx}$:$

Vedclass · Accepted Answer

$\frac{1}{A}$

Vedclass · Answer

(D) We are given the limit: $\lim _{n \rightarrow \infty} \frac{\cos x - e^{nx}}{1 - A e^{nx}}$.Divide the numerator and the denominator by $e^{nx}$:$= \lim _{n \rightarrow \infty} \frac{\frac{\cos x}{e^{nx}} - 1}{\frac{1}{e^{nx}} - A}$.Since $x > 0$,as $n \rightarrow \infty$,$e^{nx} \rightarrow \infty$.Thus,$\frac{\cos x}{e^{nx}} \rightarrow 0$ and $\frac{1}{e^{nx}} \rightarrow 0$.Substituting these values,we get:$= \frac{0 - 1}{0 - A} = \frac{-1}{-A} = \frac{1}{A}$.

If $A \neq 0$ and $x > 0$,then $\lim _{n \rightarrow \infty} \frac{\cos x - e^{nx}}{1 - A e^{nx}} = $

Similar Questions

$\lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3} = $

If $I = \lim_{x \rightarrow 0} \sin \left( \frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}} \right)$,then the limit

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} }}{{{x^2}}}$ is equal to

If $[ \cdot ]$ denotes the greatest integer function,then evaluate the limit: $\lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{[\sin x]-[\cos x]+1}{2}$

$\mathop {\lim }\limits_{x \to {0^ + }} \left\{ {{{\left( {1 + x} \right)}^{\frac{2}{x}}}} \right\}$ is equal to (where $\{.\}$ denotes the fractional part of $x$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module