lim _{x rightarrow 0} frac{4^x-9^x}{x(4^x+9^x | vedclass

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(A) Let $L = \lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$.Since the limit is of the form $\frac{0}{0}$ as $x \rightarrow 0$,we apply $L$-Hospit

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$\log \frac{2}{3}$

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(A) Let $L = \lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$.Since the limit is of the form $\frac{0}{0}$ as $x \rightarrow 0$,we apply $L$-Hospital's rule:$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(4^x-9^x)}{\frac{d}{dx}(x(4^x+9^x))}$$L = \lim _{x \rightarrow 0} \frac{4^x \log 4 - 9^x \log 9}{(4^x+9^x) + x(4^x \log 4 + 9^x \log 9)}$Substituting $x = 0$:$L = \frac{4^0 \log 4 - 9^0 \log 9}{(4^0+9^0) + 0(4^0 \log 4 + 9^0 \log 9)}$$L = \frac{\log 4 - \log 9}{1 + 1} = \frac{\log(4/9)}{2}$$L = \frac{1}{2} \log(\frac{2^2}{3^2}) = \frac{1}{2} \cdot 2 \log(\frac{2}{3}) = \log(\frac{2}{3})$.

$\lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$ is equal to

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