Question

determine the limit as x approaches zero.

[1-cos(2x)]/[4x^2]

Answer 1

\[(1-\cos(2x))\div(4x^2)\]

Answer 2

hint-\[\cos2x=1-2\sin ^{2}x\]
use this and get the answer.

Answer 3

that is a part of what the answer says but how did you realise that cos2x equals this?

Answer 4

is it an identity?

Answer 5

oh okay thanks.. hmm

Answer 6

you can also prove it..
cos(2x)=cos(x+x)
use this and expand, you will land up as cos2x=1-2sin^2x=2cos^2x-1.

Answer 7

oh okayy. So now i have -2sin^2x over 4x^2 im not sure what to do now

Answer 8

cos2x=1-2sin^2x
2sin^2x=1-cos2x.
now we have 2sin^2x over 4x^2

Answer 9

lim x tending to 0, sinx/x=1
this is a result in limits
did you know this before?

Answer 10

hmm no I didnt.. im still a bit confused with the second last part. I thought that it would be 1-1-2sin^2x over 4x^2 ? i dont know why i went wrong. sorry

Answer 11

okay..
cos2x=1-2sin^2x
add 2sin^2x on both sides..what do you get?

Answer 12

2sin^2x + cos2x = 1 ?

Answer 13

good..
now to the equation which you've obtained, subtract cos2x from both sides.
what do you get?

Answer 14

2sin^2x = 1 - cos2x ohhh

Answer 15

yes..now use the result
lim x tending to zero, sinx/x=1
and get the answer.

Answer 16

oh okay so it would be 2/4 or 1/2, but im confused why sinx/x = 1

Answer 17

when x=0, what is sinx?

Answer 18

0

Answer 19

yes..so when x=0, sinx/x=0/0
dividing a quantity sinx over x , at x=0 is undefined.
do you agree?

Answer 20

yep