mathop {lim }limits_{x to 0} frac{{{4^x} - {9 | vedclass

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(A) Let $y = \mathop {\lim }\limits_{x 	o 0} \frac{{{4^x} - {9^x}}}{{x({4^x} + {9^x})}}$. This is a $\frac{0}{0}$ indeterminate form.Applying $L'Hosp

Vedclass · Accepted Answer

$\log \left( {\frac{2}{3}} \right)$

Vedclass · Answer

(A) Let $y = \mathop {\lim }\limits_{x 	o 0} \frac{{{4^x} - {9^x}}}{{x({4^x} + {9^x})}}$. This is a $\frac{0}{0}$ indeterminate form.Applying $L'Hospital's$ rule by differentiating the numerator and denominator with respect to $x$:$y = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(4^x - 9^x)}{\frac{d}{dx}(x(4^x + 9^x))}$$y = \mathop {\lim }\limits_{x 	o 0} \frac{4^x \ln 4 - 9^x \ln 9}{(4^x + 9^x) + x(4^x \ln 4 + 9^x \ln 9)}$Substituting $x = 0$:$y = \frac{4^0 \ln 4 - 9^0 \ln 9}{(4^0 + 9^0) + 0(4^0 \ln 4 + 9^0 \ln 9)}$$y = \frac{\ln 4 - \ln 9}{1 + 1} = \frac{\ln(4/9)}{2}$$y = \frac{\ln((2/3)^2)}{2} = \frac{2 \ln(2/3)}{2} = \ln \left( \frac{2}{3} ight)$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{4^x} - {9^x}}}{{x({4^x} + {9^x})}} = $

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