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

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

Question

(A) આપણે જાણીએ છીએ કે $\mathop {\lim }\limits_{u 	o 0} \frac{e^u - 1}{u} = 1$ અને $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$.$\mathop {\

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) આપણે જાણીએ છીએ કે $\mathop {\lim }\limits_{u 	o 0} \frac{e^u - 1}{u} = 1$ અને $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$.$\mathop {\lim }\limits_{x 	o 0} \frac{e^{\sin x} - 1}{x} = \mathop {\lim }\limits_{x 	o 0} \left( \frac{e^{\sin x} - 1}{\sin x} 	imes \frac{\sin x}{x} ight)$$= \left( \mathop {\lim }\limits_{\sin x 	o 0} \frac{e^{\sin x} - 1}{\sin x} ight) 	imes \left( \mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} ight)$$= 1 	imes 1 = 1$.વૈકલ્પિક રીતે,$L$-Hospital ના નિયમનો ઉપયોગ કરતા:$\mathop {\lim }\limits_{x 	o 0} \frac{e^{\sin x} - 1}{x} = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(e^{\sin x} - 1)}{\frac{d}{dx}(x)}$$= \mathop {\lim }\limits_{x 	o 0} \frac{\cos x \cdot e^{\sin x}}{1} = \cos(0) \cdot e^{\sin(0)} = 1 \cdot e^0 = 1$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\sin x}} - 1}}{x} = $

Similar Questions

$\mathop {\lim }\limits_{x \to \infty } \frac{{\log x}}{{{x^n}}}, \; n > 0$ ની કિંમત શું છે?

$\lim _{x \rightarrow 2}(x-1)^{\frac{1}{3x-6}} = $

$\mathop {\lim }\limits_{x \to 0^+} {x^x} = $

$\mathop {\lim }\limits_{x \to \pi /2} \left[ {x\tan x - \left( {\frac{\pi }{2}} \right)\sec x} \right] = $

જો $f(a) = 2$,$f'(a) = 1$,$g(a) = -3$,$g'(a) = -1$ હોય,તો $\mathop {\lim }\limits_{x \to a} \,\frac{f(a)g(x) - f(x)g(a)}{x - a} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module