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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Aicharya · Accepted Answer

Given that, $f(2)=4$ and $f^{\prime}(2)=1$ $ \begin{aligned} 	herefore \quad & \lim _{x ightarrow 2} \frac{x f(2)-2 f(x)}{x-2} \ & =\lim _{x ightarrow 2} \frac{x f(2)-2 f(2)+2 f(2)-2 f(x)}{x-2} \ = & \lim _{x ightarrow 2} f(2)-2 \lim _{x ightarrow 2} \frac{f(x)-f(2)}{x-2} \ = & f(2)-2 f^{\prime}(2) \ = & 4-2(1) \ = & 2 \end{aligned} $

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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