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can anyone tell why dont anyone want to solve my problem if there is any problem in graphics then i can upload another pic............. please explain fully ….if it has 0/0 form …..then please tell why

can anyone tell why dont anyone want to solve my problem if there is any problem in graphics then i can upload another pic.............
please explain fully ….if it has 0/0 form …..then please tell why

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Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:

Hello Student,
Please find answer to your question below

Forget the 0/0 form if that making you damn confusing.
I am solving this by standard method. And remember L’Hospital is a shortcut method.
I am assuming not any other data is given.
And it is obvious that limit exist at x = 0
For the limit
\lim_{x\rightarrow 0}\frac{2f(x)-3f(2x)+f(4x)}{x^{2}}
to exist
f(x) = ax^{2} + c; c = constant
So now consider the function inside limit
\frac{2(ax^{2}+c)-3(4ax^{2}+c)+(16ax^{2}+c)}{x^{2}}
=6a
f'(x) = 2ax
\lim_{x\rightarrow 0}\frac{2f'(x)-6f'(2x)+4f'(4x)}{2x}
\lim_{x\rightarrow 0}\frac{2(2ax)-6(4ax)+4(8ax)}{2x}
=6a
f''(x) = 2a
f''(0) = 4
\Rightarrow a = 2
So answer is 6(2) which is 12.
Hope i clarify your doubt.

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