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Limits without L'Hopitals rule

  • Thread starter vbplaya
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  • #1
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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).
 

Answers and Replies

  • #2
Tom Mattson
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Hello, and welcome to Physics Forums. :smile:

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. :biggrin:

OK, now that you've read the notice, what have you done on this problem?
 
  • #3
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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
lurflurf
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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).
Trig is your friend
[tex]\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)}[/tex]

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
dextercioby
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Use the Taylor expansions of the trig functions around 0

[tex] \sin x\simeq x-\frac{x^{3}}{3!} [/tex]

[tex] \tan x\simeq x+\frac{x^{3}}{3} [/tex]

It should come up to [itex] -\frac{1}{2} [/itex].

Daniel.
 
  • #6
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lurflurf said:
Trig is your friend
[tex]\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)}[/tex]
i'm sorry, but i don't how u get from [tex]\frac{\sin(x)-\tan(x)}{x^3}[/tex] to [tex]-\frac{\sin(x)}{x} \ \left(\frac{\sin\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)^2 \ \frac{1}{2\cos(x)}[/tex]
 
  • #7
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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
lurflurf
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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
 
  • #9
lurflurf
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vbplaya said:
i'm sorry, but i don't how u get from [tex]\frac{\sin(x)-\tan(x)}{x^3}[/tex] to [tex]-\frac{\sin(x)}{x} \ \left(\frac{\sin\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)^2 \ \frac{1}{2\cos(x)}[/tex]
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
 

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