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?

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.

To find the limit of the function (cos(x) - 1) / x as x approaches 0, you can use algebraic manipulation and apply the limit properties.

Let's first simplify the function:

(cos(x) - 1) / x = (cos(x) - 1) * (1 / x)

Now, consider the limit of each factor separately.

The limit of (1 / x) as x approaches 0 can be found by evaluating the limit as x approaches 0 from both the left and the right:

lim(x->0+) 1/x = +∞ (approaches positive infinity)
lim(x->0-) 1/x = -∞ (approaches negative infinity)

The limit of (cos(x) - 1) as x approaches 0 can be found by evaluating the limit as x approaches 0 from both the left and the right:

lim(x->0+) (cos(x) - 1) = -1 (approaches -1)
lim(x->0-) (cos(x) - 1) = -1 (approaches -1)

Now, let's combine the limits using the limit properties.

lim(x->0+ or 0-) [(cos(x) - 1) * (1 / x)] = lim(x->0+ or 0-) (cos(x) - 1) * lim(x->0+ or 0-) (1 / x) = -1 * ±∞

The product of -1 and positive or negative infinity is an indeterminate form. In this case, the limit does not exist.

Therefore, your initial understanding that the limit is equal to 0 is incorrect. The limit of the function (cos(x) - 1) / x as x approaches 0 does not exist.