# Limits without L'Hopitals rule

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).

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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 https://www.physicsforums.com/showthread.php?t=4825 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?

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.

Homework Helper
vbplaya said:
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).
$$\frac{\sin(x)-\tan(x)}{x^3}=-\frac{\sin(x)}{x} \ \left(\frac{\sin\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)^2 \ \frac{1}{2\cos(x)}$$

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?

Homework Helper
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.

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

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

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.

Homework Helper
vbplaya said:
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.
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

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