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

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(C) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}$સ્વરૂપ $\frac{0}{0}$ હોવાથી,આપણે $L'	ext{Hospital's r

Vedclass · Accepted Answer

$2$

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(C) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}$સ્વરૂપ $\frac{0}{0}$ હોવાથી,આપણે $L'	ext{Hospital's rule}$ નો ઉપયોગ કરીશું:$= \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + {e^{ - x}} - 2}}{{1 - \cos x}}$ફરીથી $L'	ext{Hospital's rule}$ નો ઉપયોગ કરતા:$= \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$ત્રીજી વખત $L'	ext{Hospital's rule}$ નો ઉપયોગ કરતા:$= \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + {e^{ - x}}}}{{\cos x}} = \frac{{1 + 1}}{1} = 2$

લક્ષ $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}$ ની કિંમત શોધો.

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