Question

limit x -> 0
(cos(tan x) - cos x)/(x^4)

Answer 1

substituting gives 0/0 so it might help to use l'hopitals rule

Answer 2

please do...
it's a pretty long expression

Answer 3

yea you are right! lol
first pass is
-sin(tanx).sec^2x + sin x
----------------------
4x^3

Answer 4

that gives 0/0 as well!!

Answer 5

indeed.
nothing can save us at this point.

Answer 6

yes I dont think this is the way to go
the denominator would take another 2 applications before it became a cosntant

Answer 7

-cos(tanx)sec^2(x) + sin(tanx)2sec^2(x)tan(x) + cos(x)
----------------------------------------------------------------------
12x^2

hmm...

Answer 8

yea - another 0 deniminator

Answer 9

there has to be a way of simplifying the numerator first

Answer 10

yes that looks promising

Answer 11

i recall the identity
cos C - cos D = 2sin(D-C/2)sin(C+D/2)

Answer 12

that'd do
cos(tan x) - cos x

= 2sin((x-tanx)/2)sin((x+tanx)/2)

Answer 13

now use l'hopital

Answer 14

oh i thought the first cos was jsust the cos of tanx NOt cos (tanx + cosx)

Answer 15

no, it's
(cos(tan(x)) - cos(x))/(x^4)
so it really is the cos of just tan(x)

Answer 16

ah - i see Yes

Answer 17

but i think l'hopital's will still be unhelpful

Answer 18

I reckon the numerator simplifies to -2sec^2(x)sin(x) after the first differentiation using l'hopital

Answer 19

https://www.wolframalpha.com/input/?i=limit+as+x+approaches+0+of+(cos(tan+x)+-+cos+x)%2F(x%5E4)

thats what wolfram gives

Answer 20

unfortunately i have to go now
must admit i'm stuck though

Answer 21

ah.... but how do they get there? Let me keep going with what I've got and see if i get -1/3

Answer 22

still working on it. its a tough one

Answer 23

looks like they've written it as a series expansion
note that -1/3 is the first term of the exxxpansion

I guess they have use Mclaurins series

Answer 24

but how did they get -1/3 as the first term f'(0) would be indeterminate
i guess they haven't used mcLaurins

Answer 25

Ah the Taylor series

Answer 26

- i always have trouble with that!

Answer 27

I'm afraid I cant remember much about the Taylor series . I know you have to find derivatives though...

Answer 28

The first term of the Taylor Series is f(0) which is indeterminate ???

Answer 29

OK. I got it!!

If you just do the Maclaurin expansion of the numerator you get

-x^4/3 - 2x^6/15 - 43x^8/840 and so on

now if you divide this by x^4 you get

-1/3 - 2x^2/15 - 43x^4/840 - ........

so the Limit as x -> 0 is -1/3


Taylor expansion is f(0) + xf'(0) + x^2.f''(0)/2! + x^3.f'''(0)/3! +.....

Answer 30

How did you get that expansion?

Answer 31

i was going the ask the same question !

Answer 32

f(0) for the numerator is -1
f'(0) is 0
f"(0) is 0
- then finding the next derivative would be hard work...

Answer 33

@ganeshie8 - any ideas on this one?

Answer 34

i don't see how he got that series

Answer 35

the successive derivatives will be horrendously long!!

Answer 36

have to go anyway

Answer 37

You are absolutely right!
The derivatives are ridiculously long.
But you really only need to go to the 4th derivative because that's the one that has the 4th degree which gets cancelled out by the denominator.
Even so, it was very difficult to differentiate to the 4th derivative.

Answer 38

If your a bit lazy you can use an online derivative calculator for the numerator

Answer 39

But I love a challenge

Answer 40

By the way f (0) for the numerator is also 0

Answer 41

so I do all that for you @CalcSoBad and I don't even get a thanks!
I won't be helping you again, that's for sure