mathop {lim }limits_{x to {1^ + }} frac{{{{le | vedclass

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Question

(B) ધારો કે $t = \{x\}$. જેમ $x 	o 1^+$,તેમ $t 	o 0^+$.પદાવલિ $\mathop {\lim }\limits_{t 	o 0^+} \frac{{(1 + t)^{1/t} - e^{1 - t/2}}}{1 - \cos t}$

Vedclass · Accepted Answer

$\frac{{2e}}{3}$

Vedclass · Answer

(B) ધારો કે $t = \{x\}$. જેમ $x 	o 1^+$,તેમ $t 	o 0^+$.પદાવલિ $\mathop {\lim }\limits_{t 	o 0^+} \frac{{(1 + t)^{1/t} - e^{1 - t/2}}}{1 - \cos t}$ બને છે.$(1 + t)^{1/t} = e^{1 - t/2 + t^2/3 - \dots} = e^{1 - t/2} \cdot e^{t^2/3 - \dots}$ વિસ્તરણનો ઉપયોગ કરતા.લિમિટમાં કિંમત મૂકતા:$\mathop {\lim }\limits_{t 	o 0^+} \frac{e^{1 - t/2} (e^{t^2/3} - 1)}{t^2/2} = \mathop {\lim }\limits_{t 	o 0^+} \frac{e^{1 - t/2} (1 + t^2/3 + \dots - 1)}{t^2/2}$.$= \mathop {\lim }\limits_{t 	o 0^+} \frac{e^{1 - t/2} \cdot t^2/3}{t^2/2} = \frac{e}{3} \cdot 2 = \frac{2e}{3}$.

Similar Questions

જો $\alpha = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{1 - \cos x}$ અને $\beta = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{\sqrt{1+x^2} - \sqrt{1-x^2}}$,હોય તો

$\mathop {\lim }\limits_{x \to \pi /2} \frac{2x - \pi}{\cos x} = $

$\lim _{x \rightarrow 0} \left( \frac{3^{x}-1}{x} \right)$ ની કિંમત શોધો.

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$\lim _{x \rightarrow \frac{\pi}{2}} \left(\frac{2x-\pi}{\cos x}\right)$ ની કિંમત શોધો.

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