Limit of a function as x goes to 1

[nothingkwt]

f(x) = 1-cos(x-1) / (x-1)²

Can someone please explain how this can be done other than L'Hospital's rule? I tried but I just don't understand how. I know the answer is 1/2 though.

I tried separating to 2 limits but i got ∞ - ∞ and that is obviously wrong but I don't know what my mistake was.

 

[Jolb]

You could do this by doing a series expansion for the numerator and looking at the limit of the leading order expression. This is a useful series expansion:

cos(x-1) ≈ 1 - (x-1)²/2! + (x-1)⁴/4! + ...
This is basically the definition of cosine, but you can get it through taylor expansion about x=1. Note that as x goes to 1, the higher order terms are negligible.
If you plug this into the numerator of your expression, you get some nice cancellation happening.

 

[mfb]

There are brackets missing for the numerator, I think.

If you know good upper and lower estimates for cos() around 0, you can use them. The full taylor series is the better solution, if you can use that.

 

[Mark44]

A Taylor's Series expansion seems to me to be the way to go.

 

[iRaid]

A Taylor's Series expansion seems to me to be the way to go.

How would you go about even expressing that as a taylor series

 

[Office_Shredder]

iRaid, can you express cos(x-1) as a Taylor series around x=1?

Actually Jolb has it above, but it doesn't do much if you don't know why that is what it is.