lim _{x rightarrow 0}left(frac{(x+2 cos x)^{3 | vedclass

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(B) Let $f(x) = \frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}$.As $x \rightarrow 0$,$f(x) \rightarr

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$1$

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(B) Let $f(x) = \frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}$.As $x \rightarrow 0$,$f(x) \rightarrow \frac{2^3 + 2(2^2) + 3 \sin 2}{2^3 + 2(2^2) + 3 \sin 2} = 1$.This is a $1^{\infty}$ form.Using the formula $\lim_{x \rightarrow a} f(x)^{g(x)} = e^{\lim_{x \rightarrow a} g(x)(f(x)-1)}$,we get:$L = e^{\lim_{x \rightarrow 0} \frac{100}{x} \left( \frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x) - (x+2)^{3}-2(x+2)^{2}-3 \sin (x+2)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)} \right)}$.Let $h(u) = u^3 + 2u^2 + 3 \sin u$. Then $f(x) = \frac{h(x+2 \cos x)}{h(x+2)}$.As $x \rightarrow 0$,$f(x) - 1 = \frac{h(x+2 \cos x) - h(x+2)}{h(x+2)}$.Using the derivative $h'(u) = 3u^2 + 4u + 3 \cos u$,the limit becomes $e^{100 \cdot \frac{h'(2) \cdot \lim_{x \rightarrow 0} \frac{(x+2 \cos x) - (x+2)}{x}}{h(2)}}$.Since $\lim_{x \rightarrow 0} \frac{2 \cos x - 2}{x} = \lim_{x \rightarrow 0} \frac{-2 \sin x}{1} = 0$,the exponent is $0$.Thus,$L = e^0 = 1$.

$\lim _{x \rightarrow 0}\left(\frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}\right)^{\frac{100}{x}}$ is equal to $.....$

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