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Another lim problem, but different:(

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\[f(x) = \left[\begin{matrix}x^2-1 & x<1 \\ 4-x & x>=1\end{matrix}\right]\]
I can do the lim, but how do i do it on that:O it lin x->1
\[\lim_{x \rightarrow 1}\]

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Other answers:

|dw:1346015959313:dw| Your graph will look something like this
What is the limit for the function x^2 - 1 as it approaches 1?
Erm. -inf. right?
Hmm, not quite. You want to approach x = 1.
Follow the graph of x^2 -1 as it approaches x =1. What is the y-value?
I think:3
Sorry *.* i'm not very good a limits:(
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.
What would you say the limit was for that equation?
Ok, are the two limits for both equations the same?
Nope. Oh:3. Dur dur. You compare the 2 equations:3
What is your final answer for the problem?
@Denebel one last Q. It also says to find x->+ and -. In this cast, thats DNE too. right?
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.
But those are not the same for either left or right.
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
Oh so you only look at each problem separately then :O
so lim x->1- = 0 and + = 3 and x-> = DNE
<3 Thank you so much!

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