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[SOLVED] Limits |
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| Sep25-07, 08:18 PM | #1 |
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[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... |
| 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?)
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| Sep26-07, 12:22 PM | #3 |
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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)?
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| Sep26-07, 12:29 PM | #5 |
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(x-3}^2. Oh yeah, so that takes away the square root, and after everything, it leaves 0. thank you.
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| 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 ?
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| Sep26-07, 12:51 PM | #7 |
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it is 1, not 0.
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| Sep26-07, 12:55 PM | #8 |
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Well, then your book has a typo..
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| Sep26-07, 12:58 PM | #9 |
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it's not from my book, it was my teacher.
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| 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.
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| 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 |
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allright, thankyou.
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