# Limit as x approaches 0

• Apr 15th 2010, 06:56 AM
UC151CPR
Limit as x approaches 0
possibly L'Hopital & FTC
• Apr 15th 2010, 07:01 AM
General
Exactly.

If you apply L'Hospital's Rule and FTC, you will get:

$\displaystyle \frac{1}{3} \, \lim_{x\to 0} \dfrac{ \dfrac{sin(x)}{x} - 1 }{x^2}$

so ?
• Apr 15th 2010, 07:04 AM
UC151CPR
Quote:

Originally Posted by General
Exactly.

If you apply L'Hospital's Rule and FTC, you will get:

$\displaystyle \frac{1}{3} \, \lim_{x\to 0} \dfrac{ \dfrac{sin(x)}{x} - 1 }{x^2}$

so ?

Does Si = Sin?
• Apr 15th 2010, 07:11 AM
General
No.
Why did you ask that ?
By the FTC:

$\displaystyle \frac{d}{dx} \left( \int_0^x f(t) \, dt \right) = f(x)$ ..
• Apr 15th 2010, 07:21 AM
UC151CPR
Quote:

Originally Posted by General
Exactly.

If you apply L'Hospital's Rule and FTC, you will get:

$\displaystyle \frac{1}{3} \, \lim_{x\to 0} \dfrac{ \dfrac{sin(x)}{x} - 1 }{x^2}$

so ?

Do we have to do L'Hopital again? Since sinx/x =1
and then subtracting 1 gets you 0. so 0/0 ?

or is the limit just 0?
• Apr 15th 2010, 07:41 AM
General
Quote:

Originally Posted by UC151CPR
Do we have to do L'Hopital again? Since sinx/x =1
and then subtracting 1 gets you 0. so 0/0 ?

or is the limit just 0?

Its still 0/0 not 0 ..

We need to use L'Hospital's Rule two times again to evaluate the limit ..
Or we can use the infinite series to evaluate it ..
Your final answer should be $\displaystyle \frac{-1}{18}$ ..