Question

find the limit as x approaches 0,( x-sinx)/x^3

Answer 1

apply the same l'hospital rule

Answer 2

differentiating will give \[\frac{ 1-\cos x }{ 3x^{2} }\]

Answer 3

yes,apply the same l'hospital rule once again

Answer 4

substituting x=0, \[\frac{ 1-\cos0 }{ 3(0)^{2} }=\frac{ 1-1 }{ 0 }=\frac{ 0 }{ 0}=0\]

Answer 5

\(\large \lim_{x\to\ 0}\frac{(x-sinx)}{x^3}\)

\(\large =\dfrac{d}{dx}\left(\dfrac{x-sinx}{x^3}\right)\)

\(\large =\dfrac{1-cosx}{3x^2}\)

\(\large =\dfrac{d}{dx}\left(\dfrac{1-cosx}{3x^2}\right)\)

\(\large =\dfrac{-(-sinx)}{6x^1}\)

\(\large =\dfrac{d}{dx}\left(\dfrac{sinx}{6x^1}\right)\)

\(\large =\dfrac{cosx}{6}\)

\(\large =\dfrac{1}{6} (as~~ cos0=1)\)

Answer 6

how many times do i have to differentiate?

Answer 7

sorry three times here

Answer 8

how do i know that i have to end on the first,second ,third, forth etc. differential

Answer 9

for example after first time try to put x=0 then u will get 0/0 sp thats dubious,
here as the deniminator had taken three steps for eliminating it ,it took 3 here

Answer 10

okay,so when i continue getting0/0 i have to continue differentiation until i am sure of not getting 0

Answer 11

Is that it?

Answer 12

yes most of the times

Answer 13

thank you.