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[SOLVED] Limits

 
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Sep25-07, 08:18 PM   #1
 

[SOLVED] Limits


I have two...

1. The problem statement, all variables and given/known data
The the limit


2. Relevant equations
[latex]\lim_{x \rightarrow 1} \frac{1-cosx}{x^2}[/latex]


3. The attempt at a solution
I figured to just plug in 1, but I wanted to make sure....

1. The problem statement, all variables and given/known data
Find the limit


2. Relevant equations
[latex]\lim_{x \rightarrow 3} \frac{\sqrt{x^2-6x+9}}{x-3}[/latex]


3. The attempt at a solution
I plugged in the 3, and got 3/0, then I got lost...
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Sep25-07, 09:31 PM   #2
 
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In 2, did you try simplifying the numerator? (What are the roots of the polynomial?)
Sep26-07, 12:22 PM   #3
 
Yes, I tried doing that.

[latex](x-3)(x-3)[/latex]

However, I forgot how to get rid of that radical. Squaring wouldn't work, so I have no idea.


Also, no thoughts on the first one?
Sep26-07, 12:24 PM   #4
 
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[SOLVED] Limits


What is a short hand expression for (x-3)(x-3)?
Sep26-07, 12:29 PM   #5
 
(x-3}^2. Oh yeah, so that takes away the square root, and after everything, it leaves 0. thank you.
Sep26-07, 12:41 PM   #6
 
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In the first one, are you sure the problem isn't x->0 instead of x->1 ?
Sep26-07, 12:51 PM   #7
 
it is 1, not 0.
Sep26-07, 12:55 PM   #8
 
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Well, then your book has a typo..
Sep26-07, 12:58 PM   #9
 
it's not from my book, it was my teacher.
Sep26-07, 01:01 PM   #10
 
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Well, it's 99% certain that your teacher meant to write 0 instead of 1. With 1, it's trivial, since both the numerator and denominator are finite, nonzero constants in that limit.
Sep26-07, 01:01 PM   #11
 
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Then he either blundered, or tried to fool you.

Your function is defined&continuous on all values of x except x=0.

Your original approach is perfetly valid in the case of x=1.
Sep26-07, 01:06 PM   #12
 
allright, thankyou.
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