lim _{x rightarrow 2}left[left(x^2-4 x+4right | vedclass

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(D) We evaluate the limit by splitting it into two parts: $\lim _{x \rightarrow 2} \left[ (x-2)^2 \cos \left(\frac{2}{x-2}\right) \right] + \lim _{x \

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(D) We evaluate the limit by splitting it into two parts: $\lim _{x ightarrow 2} \left[ (x-2)^2 \cos \left(\frac{2}{x-2}ight) ight] + \lim _{x ightarrow 2} \left[ \frac{x^2-4}{x^3-2 x-4} ight]$ For the first part,$\lim _{x ightarrow 2} (x-2)^2 \cos \left(\frac{2}{x-2}ight)$. Since $(x-2)^2 ightarrow 0$ and $\cos \left(\frac{2}{x-2}ight)$ is bounded in $[-1, 1]$,by the Squeeze Theorem,the limit is $0 	imes [-1, 1] = 0$. For the second part,$\lim _{x ightarrow 2} \frac{(x-2)(x+2)}{(x-2)(x^2+2x+2)} = \lim _{x ightarrow 2} \frac{x+2}{x^2+2x+2}$. Substituting $x=2$,we get $\frac{2+2}{2^2+2(2)+2} = \frac{4}{4+4+2} = \frac{4}{10} = \frac{2}{5}$. Thus,the total limit is $0 + \frac{2}{5} = \frac{2}{5}$.

$\lim _{x \rightarrow 2}\left[\left(x^2-4 x+4\right) \cos \left(\frac{2}{x-2}\right)+\frac{x^2-4}{x^3-2 x-4}\right]=$

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