# Limits without L'Hopitals rule

1. Sep 25, 2005

### vbplaya

how do i find lim x→0 (sin x - tan x)/x³ without using l'hopitals rule?

also, can someone explain this to me, because I don't understand it.
given that lim x→c f(x)=L >0. Prove that there exists an open interval (a,b) containing c such that f(x) > 0 for all x ≠ c in (a,b).

2. Sep 25, 2005

### Tom Mattson

Staff Emeritus
Hello, and welcome to Physics Forums.

I've moved this to our Homework section. Please post all homework-type questions here in the future. Also, please see the notice at the top of this Forum.

I'll pause for you to read it.

OK, now that you've read the notice, what have you done on this problem?

3. Sep 25, 2005

### vbplaya

for the first one, i've tried to simplify sin x - tan x, but that didn't seem to work out.. and I don't know what else to try.
and for the second one, i've got no clue where to start.

4. Sep 26, 2005

### lurflurf

$$\frac{\sin(x)-\tan(x)}{x^3}=-\frac{\sin(x)}{x} \ \left(\frac{\sin\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)^2 \ \frac{1}{2\cos(x)}$$

for this
given that lim x→c f(x)=L >0. Prove that there exists an open interval (a,b) containing c such that f(x) > 0 for all x ≠ c in (a,b)
The existence of a limit tells us something about a certain open interval
What is it?

5. Sep 26, 2005

### dextercioby

Use the Taylor expansions of the trig functions around 0

$$\sin x\simeq x-\frac{x^{3}}{3!}$$

$$\tan x\simeq x+\frac{x^{3}}{3}$$

It should come up to $-\frac{1}{2}$.

Daniel.

6. Sep 26, 2005

### vbplaya

i'm sorry, but i don't how u get from $$\frac{\sin(x)-\tan(x)}{x^3}$$ to $$-\frac{\sin(x)}{x} \ \left(\frac{\sin\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)^2 \ \frac{1}{2\cos(x)}$$

7. Sep 26, 2005

### vbplaya

also for lim x→c f(x)= L >0. prove that there exists and open interal (a,b) containing c such that f(x)>0 for all x ≠ c in (a,b)

do I just pick out numbers and plug them in? because I don't know what else to do.

8. Sep 26, 2005

### lurflurf

If the limit exist then for any h>0 there exist an open interval such that
|f(x)-L|<h for all x in the open interval
in particular if L>0 then there exist an open interval such that
|f(x)-L|<L for all x in the open interval
or equivalently
0<f(x)<2L so that
0<f(x) on some open interval

9. Sep 26, 2005

### lurflurf

use
sin(x)-tan(x)=sin(x)(cos(x)-1)/cos(x)
cos(x)-1=cos(x)-cos(0)=-2sin((x+0)/2)sin((x-0)/2)=-2(sin(x/2))^2