trigonometry limits (need help as soo as possible)

habosh

13

I've been trying to solve this since like 4 hours ago :frown:

lim(x-->0) [1-cos5h]/[cos7h-1]
find the limit of course, :frown:

Reply 1

dvs

10

Do you mean h->0?

Anyway, for small h, cos(h)=~1-0.5h^2. So:
lim [1-cos(5h)]/[cos(7h)-1] = lim [1-1+0.5(5h)^2]/[1-0.5(7h)^2-1]
= lim 25h/49h
= 25/49

Reply 2

habosh

OP

13

dvs

Do you mean h->0?

Anyway, for small h, cos(h)=~1-0.5h^2. So:
lim [1-cos(5h)]/[cos(7h)-1] = lim [1-1+0.5(5h)^2]/[1-0.5(7h)^2-1]
= lim 25h/49h
= 25/49

I still don't get it :frown:

what is the .5 from where did you get it :confused:

Reply 3

Gaz031

15

habosh

I still don't get it :frown:

what is the .5 from where did you get it :confused:

He used the taylor expansions.

Reply 4

dvs

10

It's from the series expansion for cos(x).

You can also do that limit using the identity cos(2x)=1-2sin^2(x) and lim_{x->0} sinx/x = 1.

Reply 5

habosh

OP

13

but it's cos5h so how can I use the cos2x rule, I've tried to do the cos(2h+3h) thingie and manipulate the function but well I failed :frown:

Reply 6

BCHL85

12

see this
http://www.thestudentroom.co.uk/t174084.html

Reply 7

dvs

10

cos(5h) = 1 - 2sin^2(5h/2)
cos(7h) = 1 - 2sin^2(7h/2)

So your limit becomes:
lim [2sin^2(5h/2)]/[-2sin^2(7h/2)] = lim (5h/2)^2/(5h/2)^2 * (7h/2)^2/(7h/2)^2 * sin^2(5h/2)/sin^2(7h/2)
= lim (5/7)^2 * [sin(5h/2)/(5h/2)]^2 * [(7h/2)/sin(7h/2)]^2
= (5/7)^2 * 1 * 1
= 25/49

trigonometry limits (need help as soo as possible)

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