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.