f(x)= -x^2+2, 0≤ x <1
2, x=1
x^2, 1

Calculus1
- anonymous

f(x)= -x^2+2, 0≤ x <1
2, x=1
x^2, 1

Calculus1
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- anonymous

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- jim_thompson5910

how far did you get?

- anonymous

i think a) is 4 but I am not sure

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## More answers

- jim_thompson5910

a) is 4, nice work

- jim_thompson5910

you start on the left side of x = 2, then you slowly approach x = 2 along the graph and you'll get closer and closer to y = 4

- jim_thompson5910

how about b?

- anonymous

Idk how to do b :/

- jim_thompson5910

what is the limiting value of f(x) as x ---> 1

- jim_thompson5910

ie if you approach x = 1 from both directions, what does y get closer and closer to?

- anonymous

2?

- jim_thompson5910

no you approach x = 1
this means you get closer and closer to x = 1...but you actually never officially arrive at x = 1

- jim_thompson5910

does that make sense?

- anonymous

yes

- anonymous

what about f(1)?

- jim_thompson5910

f(1) is the y value of the point at x = 1

- jim_thompson5910

the limit is slowly approaching x = 1 (it never gets there officially, just gets closer and closer)
f(1) is actually at x = 1 (it's officially there and stopped)

- anonymous

so f(1) has a greater value

- jim_thompson5910

good, the limiting value as x ----> 1 is 1
but f(1) = 2
so you are correct

- anonymous

idk how to do c:/

- jim_thompson5910

what are some points where the limit does not exist?

- anonymous

where the holes are at, i think

- jim_thompson5910

that's a good guess, but look how the limit exists at x = 1
this is because you can approach x = 1 from either side and arrive at the same y value

- jim_thompson5910

however, the limit does NOT exist at x = 2, why is this?

- anonymous

cause the lim f(x) as x approaches 2- is 4 and the lim f(x) as x approaches 2+ is 3 so limf(x) as x approaches 2 does not exist because those two limits aren't equal

- jim_thompson5910

bingo

- jim_thompson5910

the left and right hand limits are NOT equal, so the limit does NOT exist

- jim_thompson5910

what's another point where the limit doesn't exist?

- anonymous

I am not sure :/

- jim_thompson5910

hint: it's not in the middle

- anonymous

wait so the limit f(x) as x approaches 2, does not exist, is part of my answer?

- jim_thompson5910

yes, there's another point though

- jim_thompson5910

c = 2 is one value
but there is one more value of c

- anonymous

is it c=0

- jim_thompson5910

yep, the limit does NOT exist at x = 0

- jim_thompson5910

because the left hand limit does not exist

- jim_thompson5910

both the left and right hand limits must exist and they must be equal for the overall limit to exist

- anonymous

why doesn't the left hand limit exist?

- jim_thompson5910

because the function ends at x = 0, so there is no bit to the left to start approaching from

- jim_thompson5910

the general rule is that the limit does not exist at endpoints because you can't approach from beyond the function

- anonymous

ooh okay got it:)

- jim_thompson5910

that's great

- anonymous

thank you:)

- jim_thompson5910

you're welcome

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