lim _{x rightarrow 0} frac{x^2 2^x-x^2 sin x- | vedclass

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(D) Given the limit: $\lim _{x \rightarrow 0} \frac{x^2(2^x - \sin x - 1)}{3^x(1 - \cos x) - (1 - \cos x)}$$= \lim _{x \rightarrow 0} \frac{x^2(2^x -

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$\frac{2}{\log 3}(\log 2-1)$

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(D) Given the limit: $\lim _{x \rightarrow 0} \frac{x^2(2^x - \sin x - 1)}{3^x(1 - \cos x) - (1 - \cos x)}$$= \lim _{x \rightarrow 0} \frac{x^2(2^x - \sin x - 1)}{(3^x - 1)(1 - \cos x)}$$= \lim _{x \rightarrow 0} \frac{2^x - \sin x - 1}{(3^x - 1) \frac{(1 - \cos x)}{x^2}}$Since $\lim _{x \rightarrow 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$,the expression becomes:$= \lim _{x \rightarrow 0} \frac{2(2^x - \sin x - 1)}{3^x - 1}$Applying $L$'Hospital's Rule:$= \lim _{x \rightarrow 0} \frac{2(2^x \ln 2 - \cos x)}{3^x \ln 3}$$= \frac{2(\ln 2 - 1)}{\ln 3}$$= \frac{2}{\log 3}(\log 2 - 1)$

$\lim _{x \rightarrow 0} \frac{x^2 2^x-x^2 \sin x-x^2}{3^x+\cos x-3^x \cos x-1}=$

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