Are you a MATHS SUPER TALENT ???

Question

Well if you are that maths genius on yahoo answers,
try this one out



only if you have that maths talent in you, go for it,else, dont waste your time cracking this puzzle




lim x tends to 0


Prove that : -

[ tan x - sin x ] / [ sin 3x - 3sinx ] = 0

Answer 1

for x=0 the fraction gives 0/0
according to hospital rule

take derivative of numerator and denominator

(1+tan^2 x-cosx)/ (3cos3x-3cosx)

again for x=0

(1+0-1)/(3-3)=0/0

repeat hospital
(2tanx(1+tan^2 x)+sinx)/(-9sin3x+3sinx)
for x=0 it gives 0/0
repeat hospital rule

(2(1+tan^2 x)+6tan^2 x(1+tan^2x))+cosx) / (-27cos3x+3cosx)

for x=0 it gives
(2+1)/(-27+3)

= - 3/24
=-1/8

remember cos0=1
sin0=0

Answer 2

Interestingly enough, you don't need L'Hôpital's rule to
do this problem, just some trig and some grinding!
Also, the answer is -1/8, not 0.
First, let's show that sin 3x = 3 sin x - 4 sin³x.
This can be obtained from
sin(2x+x) = sin 2x cos x + cos 2x sin x
= 2 sin x cos² x + (1-2 sin² x)sin x
= 2 sin x(1-sin² x) + (1-2 sin² x)sin x
= 3 sin x - 4 sin³x.
Voila!!
Now substitute this into the original expression
and write tan x = sin x/ cos x.
We get
((sin x/cos x) -sin x)/ -4 sin³ x
Cancelling sin x and simplifying yields,
(1 - cos x)/ (-4 sin²x cos x)
= (1- cos x)/ (-4(1- cos²x)cos x)
= -1/ (4(1 + cos x) cos x)
Now take the limit as x goes to 0 and
the answer is -1/8.
Done!!

Answer 3

The answer coming is -1/8........!!!!

Answer 4

im no genius sorry