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).

 

[quantumdude]

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.

 

[lurflurf]

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).

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

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?

 

[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.

 

[vbplaya]

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

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.

 

[lurflurf]

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

 

[lurflurf]

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