There are several ways to get at this limit.
Using L'Hopital's Rule, if f/g = 0/0, then the limit is f'/g'
In this case, -sin/1 = 0
You are correct.
Then, the cosine can be expressed as an infinite series:
1 - x^2 / 2! + x^4 / 4! - ...
so cos(x) - 1 = -x^2/2! + x^4/4! - ...
Divide that by x, and you are left with
-x/2! + x^3/4! - ...
which are all just powers of x. When x=0, they all vanish, so the limit is 0.
What is the limit of this function as x approaches 0?
cos(x) - 1 / x
From what I gather, the limit is equal to 0, since on the right, the function approaches from negative values close to zero and on the left, it approaches 0 with positive values close to zero. Would this be correct?
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