1@sandipan convert cos210x to cos 20x+1/2........and apply l'hospital 4 time
limx→0 { 16-16cos(cos10x)}/x4
evaluate the above limit....
please help...
-
UP 0 DOWN 0 1 15
15 Answers
1357Is the question correctly written?
answer coming out to be infinite.
11Ltx->0 16( 1 - cos (cos 10x))x4
Ltx->0 16 . 2 sin 2 ((cos 10x)/2)x4
Ltx->0 32 sin 2 ((cos 10x)/2)(cos 2 10x)/4 X (cos 2 10x)/4x4
8 Ltx->0 cos 2 10xx4
...??
done till here....how to proceed next?? [12][12][12]
39Sandi...
1 - cos(x) = 2sin2(x/2)
You missed out the half term.
11357How can you guys apply L'Hospital here. :O
Do you think it's of the form 0/0, think again :)
39If we put x = 0, it comes as 16 - 16cos(1)/0, so we can't apply L'Hospital :P
I think most of you are looking at the denominator only...differentiating 4 times will kill off the x. But our precious 0/0 is not available :P
1sorry for stupidity in post#8
@sandipan for #5
do you think this works\frac{sin^2((cos 10x)/2)}{(cos^2 10x)/4}=1???no from my side.
i think its is.........its true only wen angle for sin tends 0.....but in this case its half.
39Limx→0sin²(cos(10x)/2)cos²(10x)/4 = 4sin²(1/2).
The rest of sandi's expression evaluates to ∞.
1it is not in indeterminant form............
It is in the form of finite/0.
So answer is infinity......
11but the ans goes as:- 2....
i didn't get the answr in any way...
but surely "L-hospital" can not be applied here....
1even i am getting ∞(undefined) as answer ....
which source did u get the question?
39Ok I tried it on a limit calculator I found online. It's ∞. No arguing with computers.
11i have got it from a magazine.... then surely the manazine had written the sum wrongly.... i too am getting ∞ as the answr.......