Question

If f(x) be a twice differentiable function & f(0)=0,f"(0)=5, then \[\lim_{x \rightarrow 0}\frac{f(x)-4f(2x)+f(7x)}{x^2}=?\]

Answer 1

again use L-hospital's rule

Answer 2

differentiate the numerator and denominator separately and plug in the given values..

Answer 3

did u understand ? @jiteshmeghwal9

Answer 4

Gud explanation priyar....she is ryt....jiteshmeghwal follow d same procedure as described by her

Answer 5

\[\lim_{x \rightarrow 0}\frac{f'(x)-4f'(2x)+f'(7x)}{2x}\]Is this the first step?

Answer 6

almost..u forgot to apply the chain rule.. for f(2x) and f(7x)..

Answer 7

plz tell me how to apply chain rule

Answer 8

f'x-4f'(2x)*2+f'(7x)*7 is the numerator here

Answer 9

f(2x) differentiation is 2f'(2x) cos we must also differentiate the "2x" inside..ok?

Answer 10

\[\lim_{x \rightarrow 0}\frac{f(x)-8f'(2x)+7f'(7x)}{2x}\]

Answer 11

good job!

Answer 12

but u forgot to put f'(x)

Answer 13

its ok..differentiate again

Answer 14

sorry \[\lim_{x \rightarrow 0}\frac{f"(x)-16f"(2x)+49f"(7x)}{2}\]

Answer 15

correct! now put the values given..

Answer 16

\[{5-80+245 \over 2}={170\over2}={85}\] i gt the answer thx a lot :)

Answer 17

great job!!

Answer 18

u r welcome!