A) 2
B) -2
C) -4
D) 3
Correct Answer: C
Solution :
[c] We have, \[\underset{x\to 2}{\mathop{\lim }}\,\frac{xf(2)-2f(x)}{x-2}\,\,\,\,\,\,\,\,\,\left( \frac{0}{0} \right)\] By applying ?L? Hospital rule, we get \[=\underset{x\to 2}{\mathop{\lim }}\,f(2)-2f'(x)=f(2)-2f'(2)\] \[=4-2\times 4=-4.\]You need to login to perform this action.
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