Question

Evaluate the given limit

lim x->0 of
1-cos(5x)/sin(9x)

Answer 1

multiply and divide by (1+cos5x)

Answer 2

now i have 1-cos(5x)^2 / sin(9x) x (1+cos5x)

Answer 3

i am confused on how i would further simplify that.

Answer 4

1-cos^2 x = ?

Answer 5

sin^2x

Answer 6

so 1-cos(5x)^2 = sin^2 5x, isn't it ?
now multiply and divide by x

Answer 7

5 lim x->0 of \[\sin ^{2}x/\sin9x \times1-\cos(5x)\]

Answer 8

correct?

Answer 9

wouldn't it be sin^2 (5x) ??
and (1+cos 5x) ?

Answer 10

that is what i had, then i multiplied by 5x/5x so sin^2(5x) turns into one and im not sure what to do with the denominator

Answer 11

sin^2 5x doesn't turn into 1
\(\large \frac{sin^25x}{sin 9x(1+cos 5x)}= \frac{5(sin^25x)/5x}{9(sin 9x)/9x}*\frac{1}{(1+cos 5x)}\)
got this step?

Answer 12

then i multiply by the conjugate?

Answer 13

no.

Answer 14

i got it. i need to multiply each sin by the number located by it to make each one equal one since lim x->0 sinx/x=1

Answer 15

yes.

Answer 16

5sin5x/(9sin9x)(1-cos5x) = 1*1/1+cos5x

Answer 17

=1/1+cos5x

Answer 18

now i don't know where to go from here.

Answer 19

its sin^2 (5x) not sin 5x.
\(\large\frac{5(sin^25x)/5x}{9(sin 9x)/9x}*\frac{1}{(1+cos 5x)}=\frac{5sin 5x (1)}{9(1)}\frac{1}{1+cos 5x}\)
did u get this ?
now put x=0, directly

Answer 20

5/9 * 1/1+1 =5/9 * 1/2 = 5/18 ?

Answer 21

when u put x= 0
what is sin 0 ??

Answer 22

0

Answer 23

so?
wouldn't the answer be 0 then ?

Answer 24

correct because if you have 0 in the numerator than the whole equation turns to zero, correct?

Answer 25

yup

Answer 26

you end up multiplying the whole equation by zero, meaning the answer is 0?

Answer 27

yes!

Answer 28

thank-you very much! greatly appreciated, you were a great help.

Answer 29

welcome ^_^
did u understand all steps?
if doubts anywhere, just ask.

Answer 30

i will send you a message if i need any other help but it is a lot clearer to me now. thanks a lot.

Answer 31

ok :)