Find the indicated Limit

- anonymous

Find the indicated Limit

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- schrodinger

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- anonymous

##### 1 Attachment

- freckles

Have you looked at the left and right limit to x=0?

- anonymous

No, I don't understand how this specific example works

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## More answers

- freckles

x<0 means to the left of x=0
x>0 means to the right of x=0

- freckles

use the functions for x<0
and for x>0
to find the left and right limit

- freckles

|dw:1433390979523:dw|

- anonymous

Okay, so since there is the x->0 does that mean I use |-4-x| ?

- freckles

you use both functions
one to find the right limit
and one to find the left limit
then compare both outputs

- 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

| -4-0 | = 4
5(0) - 8 = -8
correct? so they are different?

- 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

and yes to your question 4 is not -8

- anonymous

wow thank you for your help!! for this one, is it 8?

##### 1 Attachment

- freckles

did you find left and right limit?

- 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

except we looked around 0 since we had x approaching 0

- 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

we are not concerned at what happens at x=1

- 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

anyways you can conclude since 0 isn't 8
the limit __________?

- anonymous

Oh alright, Yes the limit does not exist

- freckles

yep

- anonymous

thanks so much, these are just hard for me to understand

- freckles

like if you want to explain what is hard about it I can try to explain

- freckles

or like what you don't get about what I have said

- freckles

when evaluating limits we aren't concerned at what happens at x=a
but what happens around x=a

- 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

|dw:1433392350183:dw|

- freckles

|dw:1433392388087:dw|
this is where f is to the left of x=a

- freckles

notice the y values are approaches k

- freckles

|dw:1433392435150:dw|

- freckles

that is the left limit of x=a

- freckles

now to the right of x=a
you want to use the right piece

- freckles

|dw:1433392480605:dw|

- freckles

those y values are approaching m

- freckles

\[\lim_{x \rightarrow a^-}f(x)=k \\ \lim_{x \rightarrow a^+}f(x)=m\]

- 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

|dw:1433392571595:dw|
anyways i hope that is more clear

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