
#1
Sep2507, 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{1cosx}{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^26x+9}}{x3}[/latex] 3. The attempt at a solution I plugged in the 3, and got 3/0, then I got lost... 



#2
Sep2507, 09:31 PM

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In 2, did you try simplifying the numerator? (What are the roots of the polynomial?)




#3
Sep2607, 12:22 PM

P: 35

Yes, I tried doing that.
[latex](x3)(x3)[/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? 



#4
Sep2607, 12:24 PM

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[SOLVED] Limits
What is a short hand expression for (x3)(x3)?




#5
Sep2607, 12:29 PM

P: 35

(x3}^2. Oh yeah, so that takes away the square root, and after everything, it leaves 0. thank you.




#6
Sep2607, 12:41 PM

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P: 1,185

In the first one, are you sure the problem isn't x>0 instead of x>1 ?




#7
Sep2607, 12:51 PM

P: 35

it is 1, not 0.




#8
Sep2607, 12:55 PM

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Well, then your book has a typo..




#9
Sep2607, 12:58 PM

P: 35

it's not from my book, it was my teacher.




#10
Sep2607, 01:01 PM

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P: 1,185

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.




#11
Sep2607, 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. 



#12
Sep2607, 01:06 PM

P: 35

allright, thankyou.



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