mathop {lim }limits_{x to 0} left[ {frac{{{e^ | vedclass

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

Question

(C) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}$ which is of the form $\frac{0}{0}$.Using $L'Hospital's

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}$ which is of the form $\frac{0}{0}$.Using $L'Hospital's$ rule:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}({e^x} - {e^{\sin x}})}{\frac{d}{dx}(x - \sin x)} = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{\sin x}}\cos x}}{{1 - \cos x}}$Applying $L'Hospital's$ rule again:$\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - ({e^{\sin x}}(-\sin x) + \cos x({e^{\sin x}}\cos x))}}{{\sin x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + {e^{\sin x}}\sin x - {e^{\sin x}}\cos^2 x}}{{\sin x}}$Since $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$ and $\cos x 	o 1$,we can simplify the expression using Taylor series expansion for $e^x \approx 1+x+\frac{x^2}{2}$ and $\sin x \approx x - \frac{x^3}{6}$:$\mathop {\lim }\limits_{x 	o 0} \frac{(1+x+\frac{x^2}{2}) - (1+(x-\frac{x^3}{6})+\frac{(x-\frac{x^3}{6})^2}{2})}{x-(x-\frac{x^3}{6})} = \mathop {\lim }\limits_{x 	o 0} \frac{1+x+\frac{x^2}{2} - 1 - x + \frac{x^3}{6} - \frac{x^2}{2}}{\frac{x^3}{6}} = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{x^3}{6}}{\frac{x^3}{6}} = 1$.

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}} \right]$ is equal to

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(27 + x)}^{\frac{1}{3}}}} - 3}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$ equals.

$\mathop {Limit}\limits_{x \to 0} {(\cos 2x)^{3/x^2}}$ has the value equal to . . . . . . .

If $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1$,then

$\mathop {\lim }\limits_{x \to \alpha } \frac{{\sin x - \sin \alpha }}{{x - \alpha }} = $

$\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module