mathop {lim }limits_{x to 0} frac{{{2^x} - 1} | vedclass

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Question

(B) Using $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{{2^x} - 1}}

Vedclass · Accepted Answer

$\log 4$

Vedclass · Answer

(B) Using $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{{2^x} - 1}}{{{{(1 + x)}^{1/2}} - 1}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}({2^x} - 1)}}{{\frac{d}{{dx}}({{(1 + x)}^{1/2}} - 1)}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{{2^x}\ln 2}}{{\frac{1}{2}{{(1 + x)}^{ - 1/2}}}}$Substituting $x = 0$:$= \frac{{{2^0}\ln 2}}{{\frac{1}{2}{{(1 + 0)}^{ - 1/2}}}} = \frac{{1 \cdot \ln 2}}{{\frac{1}{2}}} = 2\ln 2 = \ln({2^2}) = \ln 4$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{{{{(1 + x)}^{1/2}} - 1}} = $

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