Question

this looks foreign

Consider the following function

f(x)=3x+1; x <-3
8; x=-3
2x-2; x>-3



Does lim (x->-3) f(x) exist? If so, whats the value of the limit?

Answer 1

Yes, this is ok. The left and right-hand limits as x goes to -3 will be -8. They are both the same number, so the limit exists. It doesn't matter that the actual value at f(-3) is something else, which in this case is +8.

Answer 2

All good now, @noye2 ?

Answer 3

What might be throwing you off is that the function is not continuous at x = -3. But the limit still exists.

Answer 4

@tcarroll010 yesss 2 more questions. Does that mean f(-3) exist...sorry if its a stupid question

Answer 5

Well, according to the function definition, which is a 3-part definition, which is fine, the function is explicitly:

f(-3) = 8 (not -8)

So, there's a big "jump" from what you would expect otherwise from -8 to positive 8 at just that point.

F(-3) exists and it is by definition 8

Answer 6

These do look odd, but these types of definitions for functions are perfectly valid.

Answer 7

so the value would be 8 @tcarroll010 for f(-3) still

Answer 8

If that's how the function is defined and you copied it correctly. No problem with those types of definitions, they're ok.

Answer 9

@tcarroll010 and its not continuous at x=-3 right?

Answer 10

That's exactly right.

Answer 11

@tcarroll010 is there a reason why???

Answer 12

It's basically because you can define a function any way you want as long as there is only one "y" value for every "x". Think of a simpler situation:

y = sqrt(x)

It's not allowed to have x < 0 (for real numbers) so you don't even have a domain for x < 0 unless you make up a different function for x < 0 and take all the definitions together.

Is there a real-world reason. There are probably some. If I thought hard and long, I might be able to come up with some.

Answer 13

@tcarroll010 thank u i learned alottt

Answer 14

uw! Good luck to you in all of your studies and thx for the recognition! @noye2