Question

hi good evening.Can someone explain to me how to solve the limit with trigonometric functions?pls. help me.....
how to evaluate this limit?
Limit 1-cos 4x/1-cos 2x as x approaches 0.

Answer 1

it's familiar but pls. explain it to me.

Answer 2

ok

Answer 3

if u get the lim 0/0
u use lhopitals rule

u differentiate the nominator and the denominator

u get lim x->0 4cos(2x)
then you factor out the constant
4(limx->0 cos(2x))
which is 4*1
so the answer is 4

Answer 4

hmm.. ok tnx.when do we use the trigo identities?

Answer 5

did u get it? or do u want me to show u more details?

Answer 6

pls2x show me some details

Answer 7

ok jus gimme a min to write it,

Answer 8

u understood y we used lhopitals? right?

Answer 9

no can u explain it to me?

Answer 10

ok when you evaluate a limit and you get 0/0 or infinity/infinity

you use lhopitals rule

lim x-> 0 f'(x)/g'(x)

Answer 11

so u differentiate top alone and bottom alone

Answer 12

in this question first d/dx (1-cos(4x))

Answer 13

you get 4sin (4x)
so far so good?

Answer 14

yup pls. continue

Answer 15

ok then u differentiate the denominator,
d/dx (1-cos(2x))
which is 2sin(2x)

Answer 16

so now u have 4sin(4x)/2sin(2x)

Answer 17

you can write it as 4cos2x using the identities of trigonomeric functions

sin2x = 2sinxcosx
sin 4x = 2sin(2x)cos(2x)
and u can simplify more, i hink u can do it

Answer 18

after the simplification u get 4cos2x
so limx>0 4cos(2x)
in limits u can factor out the constants so
4(lim x->0 cos2x)= 4*1 (because cos 0=1)

Answer 19

and im here if u need more help

Answer 20

tnx