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Limx->0 tan x /x

Limx->0 sec^2x ==> 0

I'm just not sure this is right...

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- Thread starter anderma8
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- #1

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Limx->0 tan x /x

Limx->0 sec^2x ==> 0

I'm just not sure this is right...

- #2

quasar987

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sec(x)=1/cos(x)

So how do you conclude that the limit of sec²(x) is 0 ?

So how do you conclude that the limit of sec²(x) is 0 ?

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- #4

quasar987

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Do you think that cos(0)=0 ? That's sin's job.

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Thanks very much for your help!

- #6

mathwonk

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but still everything relies on knowing cos(0) =1.

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Thanks for your info!

- #8

quasar987

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[tex]\frac{\tan(x)}{x}=\frac{1}{\cos(x)}\frac{\sin(x)}{x}[/tex]

, that

[tex]\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1[/tex]

, and that

[tex]\lim_{x\rightarrow a}f(x)g(x)=\lim_{x\rightarrow a}f(x)\cdot \lim_{x\rightarrow a}g(x)[/tex]

(provided the last two limits exist).

- #9

Gib Z

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- #10

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I found some articles on the web that sort of explain things, but I suspect that I'm going to have to read them a few times to really understand it!

Thanks for your help!

- #11

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No, we have not reviewed Taylor Series yet. We have just started that section. Might there be an area in which I could review and get a jump on what to expect?

THANKS!

- #12

Gib Z

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No it doesn't. I know Im being pedantic, but you really need an argument. tan of what? 5 bananas?

I don't get sin(x)/x = 1. Is this just something I should accept?

[\QUOTE] NO. Because thats not true for all values of x. What they were talking about is [itex]\lim_{x\to 0} \frac{\sin x}{x} =1[/itex]

I understand sin^2+cos^2=1. [\QUOTE]

Good

Your Welcome.Thanks for your help!

Just search Taylor Series In Google and Wikipedia.

- #13

Gib Z

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Quotes thing doesn't seem to want to work for me, But you get the point.

- #14

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As for point 2: limx->0 sin(x)/x = 1 vs. sin(x)/x = 1 ARE 2 different statements. Good point and I need to keep this in mind since they say very different things.

I'll search now in google for 'Taylor Series'. IK Wikipedia has also been a source of good info for me too!

Thanks!

- #15

quasar987

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I don't get sin(x)/x = 1. Is this just something I should accept?

Here is a proof: http://people.hofstra.edu/faculty/Stefan_Waner/trig/triglim.html [Broken]

If the inequality arguments are a little over your head, just remember the result because it's very useful. And come back to the proof to get your revenge when you've become stronger.

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