mathop {lim }limits_{x to 0} frac{{{{(1 + x)} | vedclass

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Question

(A) આપણે જાણીએ છીએ કે $(1 + x)^{1/x} = e^{\frac{1}{x} \ln(1 + x)}$.$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ વિસ્તરણનો ઉપયોગ કરતા:$(1 +

Vedclass · Accepted Answer

$\frac{11e}{24}$

Vedclass · Answer

(A) આપણે જાણીએ છીએ કે $(1 + x)^{1/x} = e^{\frac{1}{x} \ln(1 + x)}$.$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ વિસ્તરણનો ઉપયોગ કરતા:$(1 + x)^{1/x} = e^{\frac{1}{x}(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots)} = e^{1 - \frac{x}{2} + \frac{x^2}{3} - \dots} = e \cdot e^{-\frac{x}{2} + \frac{x^2}{3} - \dots}$.$e^u = 1 + u + \frac{u^2}{2!} + \dots$ વિસ્તરણનો ઉપયોગ કરતા,જ્યાં $u = -\frac{x}{2} + \frac{x^2}{3} - \dots$:$(1 + x)^{1/x} = e \left[ 1 + (-\frac{x}{2} + \frac{x^2}{3}) + \frac{1}{2}(-\frac{x}{2})^2 + O(x^3) ight]$$= e \left[ 1 - \frac{x}{2} + \frac{11x^2}{24} + O(x^3) ight]$.આ કિંમત લક્ષમાં મૂકતા:$\mathop {\lim }\limits_{x 	o 0} \frac{e(1 - \frac{x}{2} + \frac{11x^2}{24}) - e + \frac{ex}{2}}{x^2} = \frac{11e}{24}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/x}} - e + \frac{1}{2}ex}}{{{x^2}}}$ ની કિંમત શોધો.

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