[SOLVED] Limits


by carbz
Tags: limits, solved
carbz
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#1
Sep25-07, 08:18 PM
P: 35
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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EnumaElish
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#2
Sep25-07, 09:31 PM
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In 2, did you try simplifying the numerator? (What are the roots of the polynomial?)
carbz
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#3
Sep26-07, 12:22 PM
P: 35
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?

EnumaElish
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#4
Sep26-07, 12:24 PM
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[SOLVED] Limits


What is a short hand expression for (x-3)(x-3)?
carbz
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#5
Sep26-07, 12:29 PM
P: 35
(x-3}^2. Oh yeah, so that takes away the square root, and after everything, it leaves 0. thank you.
Avodyne
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#6
Sep26-07, 12:41 PM
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In the first one, are you sure the problem isn't x->0 instead of x->1 ?
carbz
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#7
Sep26-07, 12:51 PM
P: 35
it is 1, not 0.
arildno
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#8
Sep26-07, 12:55 PM
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Well, then your book has a typo..
carbz
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#9
Sep26-07, 12:58 PM
P: 35
it's not from my book, it was my teacher.
Avodyne
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#10
Sep26-07, 01:01 PM
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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.
arildno
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#11
Sep26-07, 01:01 PM
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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.
carbz
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#12
Sep26-07, 01:06 PM
P: 35
allright, thankyou.


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