lim (arccsinx)(cotx)
x-->0+
How do I solve this limit using L'Hopital's rule?
arcsin(x) cotx = arcsin(x)/tanx
you have 0/0, so take derivatives
1/√(1-x^2) / sec^2x → 1/1 = 1
Thank you!
To solve this limit using L'Hopital's rule, we need to take the derivative of both the numerator and the denominator until the limit is in an indeterminate form (0/0 or infinity/infinity). Let's break it down step by step:
First, let's simplify the expression:
lim(x→0+) (arccsinx)(cotx)
We can rewrite cot(x) as cos(x)/sin(x):
lim(x→0+) (arccsinx)(cosx/sinx)
We can also rewrite sin(arcsin(x)) as x:
lim(x→0+) (x)(cosx/sinx)
Now, let's take the derivative of both the numerator and the denominator:
First, the numerator:
The derivative of x with respect to x is 1.
Next, the denominator:
The derivative of cos(x) with respect to x is -sin(x).
The derivative of sin(x) with respect to x is cos(x).
Now, let's substitute the derivatives into the limit expression:
lim(x→0+) (1)(cosx)/sinx
Simplifying the expression gives:
lim(x→0+) cosx/sinx
Now, we can apply L'Hopital's rule again because the limit is still in an indeterminate form. Taking the derivatives of the numerator and the denominator:
For the numerator:
The derivative of cos(x) with respect to x is -sin(x).
For the denominator:
The derivative of sin(x) with respect to x is cos(x).
Now, let's substitute the derivatives into the limit expression:
lim(x→0+) -sinx/cosx
Simplifying the expression gives:
lim(x→0+) -tanx
Now, we can find the limit of -tanx as x approaches 0+:
lim(x→0+) -tanx = -tan(0) = 0
So, the solution to the given limit using L'Hopital's rule is 0.