Please help me with this question:
lim __1_-_cosx__
x-0 sinx
Thank you!
I will assume you mean
lim (1 - cosx)/sinx , as x ---> 0
= lim (1 - cosx)/sinx * (1+cosx)/(1+cosx)
= lim (1 - cos^2 x)/(sinx(1+cosx)
= lim sin^2 x/(sinx(1+cosx) , as x ---> 0
= lim sinx/(1+cosx) , as x ---> 0
= 0/2
= 0
Yes that was the question. Sorry I messed up the structure.
Thanks for your help!
To find the limit of the given expression, you can use L'Hôpital's rule. This rule applies when you have a limit of the form 0/0 or ∞/∞.
Here's how you can use L'Hôpital's rule to solve the limit:
1. Taking the derivative of the numerator and denominator:
- Derivative of 1 - cos(x) is 0 + sin(x) = sin(x)
- Derivative of sin(x) is cos(x)
2. Now, substitute the derivatives back into the limit expression:
lim[x->0] (sin(x))/sin(x)
3. Notice that the sin(x) terms in the numerator and denominator cancel out, leaving:
lim[x->0] 1
Therefore, the limit of the given expression as x approaches 0 is 1.