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July 26th, 2010, 10:01 PM   #1
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Another Beginner Problem I'm confused with

I'm confused with another epicly easy problem.
This one is:

lim (sinx / x)
x -> 0

I got the limit to be:
0.0174532925
I don't think that's right. I mean, It's super long O.O
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July 26th, 2010, 10:57 PM   #2
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Re: Another Beginner Problem I'm confused with

Do you know L'H˘pital's Rule?

It says that if

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July 27th, 2010, 12:30 AM   #3
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Re: Another Beginner Problem I'm confused with

So the limit is one?

Just wondering,
if I were to use a table to find this limit, how would I do that?
When I tried to use a table of values, as x approaches 0 from the left and right, I don't see 1. I'm familiar with the last 2 parts of that formula you posted, but I don't understand why it doesn't match up with my chart below.

x ...................f(x)
-0.1 ...............0.0174532837
-0.001 .............0.0174532925
-0.0001 .......... 0.0174532925
0.001 .............0.0174532925
0.01 ...............0.0174532924
0.1 .................0.0174532837
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July 27th, 2010, 01:41 AM   #4
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Re: Another Beginner Problem I'm confused with

Yes, the limit is one.

Your calculator is (apparently) in degree mode. Switch to radian mode.
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July 27th, 2010, 01:41 PM   #5
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Re: Another Beginner Problem I'm confused with

Thanks.

I have another question.

What are the differences between closed and open circle limits?

I can't see how a closed circle point can be a limit when there's nothing to limit. If it's closed, then it's continuous. Aren't limits a break in something continuous?
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July 27th, 2010, 02:26 PM   #6
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Re: Another Beginner Problem I'm confused with

An open circle means that the function is not defined at that point. For example, the point at x = 2 on the graph of (x - 2)/(x^2 - x - 2) is open because the function is not defined at x = 2.

If the circle is closed then the function is defined at that point. For example 1/(x + 1) is defined at x = 2 and therefore the circle at x = 2 is closed. The limit there, as x goes to 2, exists and is 1/3. Limits do exist at points that are on a continuous section of curve. They do not exist where the curve has a vertical asymptote, such as
(x - 2)/(x^2 - x - 2) at x = -1.
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July 28th, 2010, 02:51 AM   #7
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Hi greg1313, how do you know that the derivative of sin(x) is cos(x)? Someone who is just learning about limits hasn't yet studied exponential and trigonometric functions in depth, and proving it from first principles is likely to use knowledge of the limit you were trying to find!
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July 28th, 2010, 03:42 AM   #8
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Re: Another Beginner Problem I'm confused with

A proof of the limit of sin(x)/x as x approaches zero and the derivative of sin(x) can be found here.
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