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If f^{\prime \prime}(x) is continuous at x=0, and f^{\prime \prime}(0)=4, then the value of  \lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^2} is

Option: 1

4


Option: 2

8


Option: 3

12


Option: 4

16


Answers (1)

best_answer

\lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^2}        \quad f^{\prime \prime}(0)=4 . \\

 Required limit \lim _{x \rightarrow 0} \frac{2 f(x)-6 f^{\prime}(2 x)+4 f^{\prime \prime}\left(x_1 n\right)}{2 x} \\

\lim _{x \rightarrow 0} \frac{2 f^{\prime \prime}(x)-12 f^{\prime \prime}(2 x)+16 f^{\prime \prime}(4 x)}{2} \\

 =12 .

 

 

Posted by

himanshu.meshram

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