Differentiation Question 286
Question: If $ f(2)=4 $ , $ f’(2)=1 $ then $ \underset{x\to 2}{\mathop{\lim }}\frac{xf(2)-2f(x)}{x-2}= $
[RPET 1995, 2000]
Options:
A) 1
B) 2
C) 3
D) -2
Show Answer
Answer:
Correct Answer: B
Solution:
Given $ f(2)=4,\ \ f’(2)=1 $
$ \therefore \underset{x\to 2}{\mathop{\lim }}\frac{xf(2)-2f(x)}{x-2}=\underset{x\to 2}{\mathop{\lim }}\frac{xf(2)-2f(2)+2f(2)-2f(x)}{x-2} $
$ =\underset{x\to 2}{\mathop{\lim }}\frac{(x-2)f(2)}{x-2}-\underset{x\to 2}{\mathop{\lim }}\frac{2f(x)-2f(2)}{x-2} $
$ =f(2)-2\underset{x\to 2}{\mathop{\lim }}\frac{f(x)-f(2)}{x-2}=f(2)-2f’(2)=4-2(1)=4-2=2 $
Aliter: Applying L-Hospital rule, we get $ \underset{x\to 2}{\mathop{\lim }}\frac{f(2)-2f’(2)}{1}=2 $
BETA
