# Limits without L'Hopitals rule

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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Tom Mattson
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I'll pause for you to read it. OK, now that you've read the notice, what have you done on this problem?

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
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?

dextercioby
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.

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

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
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

lurflurf
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