Hello, I need help with this limit:
Lim ((arctanx-x)/x^3)
X->0
I can't use L'Hospital Rule or series.
I don't get what to do.
Can you give an idea?
L'Hospital's Rule:
(1/(1+x^2) - 1)/(3x^2)
= (-x^2/(1+x^2))/(3x^2)
= -1/(3(1+x^2))
-> -1/3
Well I cannot use L´Hopital´s Rule. :c
Hello! I can definitely help you find the limit of the given function without using L'Hospital's Rule or series expansion method.
To find the limit of the function ((arctanx-x)/x^3) as x approaches 0, we can use a technique called the Squeeze Theorem.
The Squeeze Theorem states that if we have three functions, f(x), g(x), and h(x), such that f(x) ≤ g(x) ≤ h(x) for all x within a certain interval, and lim f(x) = Lim h(x) = L as x approaches a common value, then g(x) also has limit value of L as x approaches the same value.
In this case, we can establish the following inequalities:
-1/x^3 ≤ ((arctanx-x)/x^3) ≤ 1/x^3
The idea is to find two functions, f(x) and h(x), whose limits we know as x approaches 0, and that sandwich the given function between them.
Let's start by evaluating the limits of f(x) = -1/x^3 and h(x) = 1/x^3 as x approaches 0:
lim (-1/x^3) as x approaches 0 = -∞ (negative infinity)
lim (1/x^3) as x approaches 0 = ∞ (positive infinity)
Since we have established that -1/x^3 ≤ ((arctanx-x)/x^3) ≤ 1/x^3, and the limits of the lower and upper bounds both go to infinity, we can conclude that the limit of the given function is also infinity as x approaches 0.
Therefore, the limit:
Lim ((arctanx-x)/x^3) as x approaches 0 = ∞
This can be verified with a graphing calculator or by substituting some values of x close to 0.
I hope this explanation helps you understand how to approach finding the limit without using L'Hospital's Rule or series expansion. If you have any further questions, feel free to ask!