If f(2)=4 and f^{prime}(2)=1,then lim _{x rig | vedclass

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(C) Given that,$f(2)=4$ and $f^{\prime}(2)=1$. We need to evaluate the limit: $L = \lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$ Adding and subtr

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$2$

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(C) Given that,$f(2)=4$ and $f^{\prime}(2)=1$. We need to evaluate the limit: $L = \lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$ Adding and subtracting $2f(2)$ in the numerator: $L = \lim _{x \rightarrow 2} \frac{x f(2)-2 f(2)+2 f(2)-2 f(x)}{x-2}$ $L = \lim _{x \rightarrow 2} \left[ \frac{f(2)(x-2)}{x-2} - 2 \frac{f(x)-f(2)}{x-2} \right]$ $L = f(2) - 2 \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}$ $L = f(2) - 2 f^{\prime}(2)$ Substituting the given values: $L = 4 - 2(1) = 4 - 2 = 2$.

If $f(2)=4$ and $f^{\prime}(2)=1$,then $\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$ is equal to

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