hawkfalcon
Another lim problem, but different:(
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hawkfalcon
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\[f(x) = \left[\begin{matrix}x^2-1 & x<1 \\ 4-x & x>=1\end{matrix}\right]\]
hawkfalcon
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I can do the lim, but how do i do it on that:O
it lin x->1
hawkfalcon
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\[\lim_{x \rightarrow 1}\]
Denebel
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|dw:1346015959313:dw|
Your graph will look something like this
Denebel
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What is the limit for the function x^2 - 1 as it approaches 1?
hawkfalcon
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Erm. -inf. right?
Denebel
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Hmm, not quite. You want to approach x = 1.
Denebel
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Follow the graph of x^2 -1 as it approaches x =1. What is the y-value?
hawkfalcon
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0
hawkfalcon
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I think:3
hawkfalcon
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Sorry *.* i'm not very good a limits:(
Denebel
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Yes :) then follow the graph of 4-x as it approaches x=1. You start from the right hand side and move back till you reach x=1.
Denebel
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What would you say the limit was for that equation?
hawkfalcon
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3
Denebel
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Ok, are the two limits for both equations the same?
hawkfalcon
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Nope. Oh:3. Dur dur. You compare the 2 equations:3
Denebel
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What is your final answer for the problem?
hawkfalcon
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DNE
hawkfalcon
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@Denebel one last Q. It also says to find x->+ and -. In this cast, thats DNE too. right?
Denebel
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Nope, as x->+ means when x approaches a number (in this case, 1) from the right hand side.
x->- means when x approaches (1) from the left hand side.
hawkfalcon
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But those are not the same for either left or right.
Denebel
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You only need to make sure they are the same when they don't give you a +/-
Since the next part of the problem asks you to state the limit when it comes from +, then when it comes from - you say what you told me before.
So when x-> 1+ is when it approaches x=1 from the right side. You only look at the function 4-x in this case.
x-> 1- is when it approaches x=1 from the left side, so look at the function x^2 - 1
hawkfalcon
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Oh so you only look at each problem separately then :O
hawkfalcon
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so lim x->1- = 0 and + = 3 and x-> = DNE
Denebel
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Yup.
hawkfalcon
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<3 Thank you so much!