Let f(x) = 4 and f'(x) = 4, then _ x 2 , xf(2) - 2f(x) x - 2 equa… - Vidyadip

MCQ

Let $f(x) = 4$ and $f'(x) = 4$, then $\mathop {\lim }\limits_{x \to 2} \,\frac{{xf(2) - 2f(x)}}{{x - 2}}$ equals

A
$2$
B
$-2$
✓
$-4$
D
$3$

✓

Answer

Correct option: C.

$-4$

c
(c) $y = \mathop {\lim }\limits_{x \to 2} \frac{{xf(2) - 2f(x)}}{{x - 2}}$

==> $y = \mathop {\lim }\limits_{x \to 2} \frac{{xf(2) - 2f(2) + 2f(2) - 2f(x)}}{{x - 2}}$

==> $y = \mathop {\lim }\limits_{x \to 2} \frac{{ - 2f(x) + 2f(2) + xf(2) - 2f(2)}}{{(x - 2)}}$

==> $y = \mathop {\lim }\limits_{x \to 2} - 2\frac{{[f(x) - f(2)]}}{{x - 2}} + \mathop {\lim }\limits_{x \to 2} \frac{{f(2).(x - 2)}}{{(x - 2)}}$

==> $y = - 2\mathop {\lim }\limits_{x \to 2} \frac{{f(x) - f(2)}}{{x - 2}} + f(2)$

==> $y = - 2\,\,\mathop {\lim }\limits_{x \to 2} f'(x) + f(2) = - \,8 + 4 = - \,4$.

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