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

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(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 :

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