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

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Question

(D) ધારો કે $y = \mathop {\lim }\limits_{x 	o 0} \frac{{x{e^x} - \log (1 + x)}}{{{x^2}}}$.આ $\frac{0}{0}$ સ્વરૂપ હોવાથી,આપણે $L'	ext{Hospital's rule

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(D) ધારો કે $y = \mathop {\lim }\limits_{x 	o 0} \frac{{x{e^x} - \log (1 + x)}}{{{x^2}}}$.આ $\frac{0}{0}$ સ્વરૂપ હોવાથી,આપણે $L'	ext{Hospital's rule}$ નો ઉપયોગ કરીશું.$y = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}(x{e^x} - \log (1 + x))}}{{\frac{d}{{dx}}({x^2})}}$$y = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + x{e^x} - \frac{1}{{1 + x}}}}{{2x}}$.ફરીથી,આ $\frac{0}{0}$ સ્વરૂપ છે,તેથી ફરીથી $L'	ext{Hospital's rule}$ નો ઉપયોગ કરતા:$y = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}({e^x} + x{e^x} - \frac{1}{{1 + x}})}}{{\frac{d}{{dx}}(2x)}}$$y = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + {e^x} + x{e^x} + \frac{1}{{{{(1 + x)}^2}}}}}{2}$$x = 0$ મૂકતા:$y = \frac{{{e^0} + {e^0} + 0 \cdot {e^0} + \frac{1}{{{{(1 + 0)}^2}}}}}{2} = \frac{{1 + 1 + 0 + 1}}{2} = \frac{3}{2}$.

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

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