- #1

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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

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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

Tom Mattson

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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 notice 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

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

- #4

lurflurf

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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

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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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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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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

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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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