anonymous
  • anonymous
Find the indicated Limit
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
freckles
  • freckles
Have you looked at the left and right limit to x=0?
anonymous
  • anonymous
No, I don't understand how this specific example works

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freckles
  • freckles
x<0 means to the left of x=0 x>0 means to the right of x=0
freckles
  • freckles
use the functions for x<0 and for x>0 to find the left and right limit
freckles
  • freckles
|dw:1433390979523:dw|
anonymous
  • anonymous
Okay, so since there is the x->0 does that mean I use |-4-x| ?
freckles
  • freckles
you use both functions one to find the right limit and one to find the left limit then compare both outputs
freckles
  • freckles
if both outputs are the same then that is what the actual limit is if both outputs are different then the limit does not exist
anonymous
  • anonymous
| -4-0 | = 4 5(0) - 8 = -8 correct? so they are different?
freckles
  • freckles
\[\lim_{x \rightarrow 0^-}f(x)=5(0)-8 \text{ here I knew to use } 5x-8 \text{ since we have } x<0 \\ \\ \lim_{x \rightarrow 0^+}f(x)=|-4-0|=4 \text{ I knew to use } |-4-x| \text{ since we have } x>0 \\ \lim_{x \rightarrow 0}f(x) \text{ doesn't exist since left not equal right}\]
freckles
  • freckles
and yes to your question 4 is not -8
anonymous
  • anonymous
wow thank you for your help!! for this one, is it 8?
freckles
  • freckles
did you find left and right limit?
freckles
  • freckles
remember we need to look at x->1+ and x->1- x->1+ means look at x>1 x->1- means look at x<1 just as we did before
freckles
  • freckles
except we looked around 0 since we had x approaching 0
anonymous
  • anonymous
1-1 = 0 x = 1 = 8 1 + 7 = 8 okay I did them in the order given, but I don't think i did the first one right?
freckles
  • freckles
we are not concerned at what happens at x=1
freckles
  • freckles
\[\lim_{x \rightarrow 1^-}f(x) \\ \text{ this mean we want to know } f(x) \\ \text{ when} x \text{ approaches } 1 \text { from left }\] the function that occurs directly to the left of the vertical line x=1 is f(x)=1-x and we know this because your function tells us to use this function when x<1 so \[\lim_{x \rightarrow 1^-}f(x)=1-1 \] and the the right function to the vertical line x=1 is f(x)=x+7 we know this because your piece-wise function says let's use this function when x>1 \[\lim_{x \rightarrow 1^+}f(x)=1+7\] now it looks like you try to find f(1) which is 8 you wouldn't say x=1=8 because that can't be true but this is unnecessary to the problem
freckles
  • freckles
anyways you can conclude since 0 isn't 8 the limit __________?
anonymous
  • anonymous
Oh alright, Yes the limit does not exist
freckles
  • freckles
yep
anonymous
  • anonymous
thanks so much, these are just hard for me to understand
freckles
  • freckles
like if you want to explain what is hard about it I can try to explain
freckles
  • freckles
or like what you don't get about what I have said
freckles
  • freckles
when evaluating limits we aren't concerned at what happens at x=a but what happens around x=a
freckles
  • freckles
\[\lim_{x \rightarrow a^-}f(x)\] you see the negative sign that looks like an exponent that tells you want to look directly to the left of x=a and see where the y's are going as x approaches a (From the left)
freckles
  • freckles
|dw:1433392350183:dw|
freckles
  • freckles
|dw:1433392388087:dw| this is where f is to the left of x=a
freckles
  • freckles
notice the y values are approaches k
freckles
  • freckles
|dw:1433392435150:dw|
freckles
  • freckles
that is the left limit of x=a
freckles
  • freckles
now to the right of x=a you want to use the right piece
freckles
  • freckles
|dw:1433392480605:dw|
freckles
  • freckles
those y values are approaching m
freckles
  • freckles
\[\lim_{x \rightarrow a^-}f(x)=k \\ \lim_{x \rightarrow a^+}f(x)=m\]
freckles
  • freckles
|dw:1433392533628:dw| if you wanted to know what the left and right limits of x=b were you will see that they are both c
freckles
  • freckles
|dw:1433392571595:dw| anyways i hope that is more clear

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