- #1

vbplaya

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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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- Thread starter vbplaya
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- #1

vbplaya

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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

quantumdude

Staff Emeritus

Science Advisor

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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?

- #3

vbplaya

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and for the second one, I've got no clue where to start.

- #4

lurflurf

Homework Helper

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Trig is your friendvbplaya said:

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).

[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

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[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

vbplaya

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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

vbplaya

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

- #8

lurflurf

Homework Helper

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If the limit exist then for any h>0 there exist an open interval such thatvbplaya said:

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

|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

Homework Helper

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usevbplaya 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]

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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