# Limit Question

• Sep 21st 2009, 04:09 PM
Power of One
Limit Question

Problem:

Lim [ cos(π+Δx) +1 ] / Δx
Δx -> 0

Hint: cos(x+y)=cosx cosy-sinx siny

My attempt:

[cosπ cosΔx + sinπ sinΔx + 1]/ Δx

[-1 cosΔx + 0 sinΔx +1 ]/ Δx

(-cosΔx + 1)/ Δx

I don't know what to do from here. I want to get rid of Δx from the denominator, but I don't know how to. Can someone please help me?
• Sep 21st 2009, 04:20 PM
hjortur
Use l'Hopitals?

$\displaystyle \lim\limits_{x\to0} \frac{-cos(x)+1}{x}=\lim\limits_{x\to0} \frac{sin(x)}{1} = 0$

Is that what you mean?
• Sep 21st 2009, 05:50 PM
Power of One
I wasn't taught that rule yet, so I doubt that is the way she wanted us to solve the problem. Can you please explain what you did to get that answer though?
• Sep 21st 2009, 07:05 PM
Prove It
Quote:

Originally Posted by Power of One
I wasn't taught that rule yet, so I doubt that is the way she wanted us to solve the problem. Can you please explain what you did to get that answer though?

L'Hospital's Rule states that if a limit is of indeterminate form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$, then

$\displaystyle \lim_{x \to 0}\frac{f(x)}{g(x)} = \lim_{x \to 0}\frac{f'(x)}{g'(x)}$.

$\displaystyle \lim_{\Delta x \to 0}\frac{-\cos{\Delta x} + 1}{\Delta x} = \lim_{\Delta x \to 0}\frac{\frac{d}{d\Delta x}(- \cos{\Delta x} + 1)}{\frac{d}{d\Delta x}(\Delta x)}$

$\displaystyle = \lim_{\Delta x \to 0}\frac{\sin{\Delta x}}{1}$

$\displaystyle = 0$

As it is, you don't need to resort to L'Hospital's Rule, as the original limit can also be solved using the sandwich theorem. Use Google to find some more info.
• Sep 21st 2009, 07:14 PM
JG89
sin(x)/1 goes to 0 as x goes to 0.
• Sep 21st 2009, 07:24 PM
Prove It
So it does. I had the famous limit $\displaystyle \lim_{x \to 0}\frac{\sin{x}}{x}$ in my head at the time...