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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
Options:
$$
\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}
$$
is equal to
Solution:
1909 Upvotes
Verified Answer
The correct answer is:
$2$
Given that, $f(2)=4$ and $f^{\prime}(2)=1$
$$
\begin{aligned}
\therefore \quad & \lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2} \\
& =\lim _{x \rightarrow 2} \frac{x f(2)-2 f(2)+2 f(2)-2 f(x)}{x-2} \\
= & \lim _{x \rightarrow 2} f(2)-2 \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \\
= & f(2)-2 f^{\prime}(2) \\
= & 4-2(1) \\
= & 2
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad & \lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2} \\
& =\lim _{x \rightarrow 2} \frac{x f(2)-2 f(2)+2 f(2)-2 f(x)}{x-2} \\
= & \lim _{x \rightarrow 2} f(2)-2 \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \\
= & f(2)-2 f^{\prime}(2) \\
= & 4-2(1) \\
= & 2
\end{aligned}
$$
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