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July 31, 2014

July 31, 2014

Posted by **Mishaka** on Saturday, October 8, 2011 at 5:59pm.

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?

- Calculus -
**Steve**, Saturday, October 8, 2011 at 6:25pmThere 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.

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