mathop {Lim}limits_{x to infty } frac{2 + 2x | vedclass

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

Question

(D) We evaluate the limit: $\mathop {Lim}\limits_{x 	o \infty } \frac{2 + 2x + \sin 2x}{(2x + \sin 2x)e^{\sin x}}$.Divide the numerator and denominat

Vedclass · Accepted Answer

non-existent

Vedclass · Answer

(D) We evaluate the limit: $\mathop {Lim}\limits_{x 	o \infty } \frac{2 + 2x + \sin 2x}{(2x + \sin 2x)e^{\sin x}}$.Divide the numerator and denominator by $x$:$\mathop {Lim}\limits_{x 	o \infty } \frac{\frac{2}{x} + 2 + \frac{\sin 2x}{x}}{(2 + \frac{\sin 2x}{x})e^{\sin x}}$.As $x 	o \infty$,$\frac{2}{x} 	o 0$ and $\frac{\sin 2x}{x} 	o 0$.The expression simplifies to $\mathop {Lim}\limits_{x 	o \infty } \frac{2}{2e^{\sin x}} = \frac{1}{e^{\sin x}}$.Since $\sin x$ oscillates between $-1$ and $1$ as $x 	o \infty$,the value of $e^{\sin x}$ oscillates between $e^{-1}$ and $e^1$.Therefore,the limit does not exist.

$\mathop {Lim}\limits_{x \to \infty } \frac{2 + 2x + \sin 2x}{(2x + \sin 2x)e^{\sin x}}$ is :

Similar Questions

The value of $\lim\limits_{n \rightarrow \infty} 6 \tan \left\{\sum\limits_{r=1}^{n} \tan ^{-1}\left(\frac{1}{r^{2}+3 r+3}\right)\right\}$ is equal to

$\mathop {\lim }\limits_{x \to a} \frac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }} = \dots$ (where $a \ne 0$)

If $\mathop {\lim }\limits_{n \to \infty } \frac{1 - 10^n}{1 + 10^{n+1}} = \frac{-\alpha}{10}$,then find the value of $\alpha$.

$\lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module