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

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(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 {\

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(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

લક્ષની કિંમત શોધો: $\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}$

જ્યારે $x \rightarrow 0$ હોય ત્યારે $\left\{\frac{1}{x} \sqrt{1+x}-\sqrt{1+\frac{1}{x^{2}}}\right\}$ ની લક્ષ કિંમત:

$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\sin \left( n \right)}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} \right)} \right]^n}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}} \right]$ ની કિંમત શોધો.

જો $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$ હોય,તો $\mathop {\lim }\limits_{\alpha \to 0} \frac{f(1 - \alpha) - f(1)}{\alpha^3 + 3\alpha}$ ની કિંમત શોધો.

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