# Limit of a function as x goes to 1

• nothingkwt

#### 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.

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.

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.

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

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

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.