Question

How do I prove that the function y=sqrt(2x+3) is only continuous past -1.5 with limits?

Answer 1

sqroot should not be negative

Answer 2

I need to prove it with limits

Answer 3

take the limit of that function at x=-3/2
from both sides! if the limit exist and have a value then the function must be continues past -1.5

Answer 4

So I have to plug in numbers really close to -1.5 on both sides?

Answer 5

to show what im saying here
\[\lim_{x \rightarrow \frac{-3^-}{2}}\sqrt{2x+3}=\lim_{x \rightarrow \frac{-3^+}{2}}\sqrt{2x+3}\]

Answer 6

we must show that this exist and true!

Answer 7

well that's one way of thinking about it^_^

Answer 8

Would I say that the limit approaching from the left is undefined or DNE?

Answer 9

open study is going crazy lol

Answer 10

what?

Answer 11

well check the limit and see! why would it be undefined?

Answer 12

to me it is tending to some value?

Answer 13

Because the function doesn't exist before (-1.5,0)

Answer 14

but it is going towards some value even if there nothing before that point

Answer 15

So I do not have to take the left hand limit because the function does not exist past -1.5?

Answer 16

yeah! the function needs not the part from the left!

Answer 17

Ok. So I only take the right limit?

Answer 18

yes! your function will be continuos from x>=-3/2 other than that
the function doesn't exist

Answer 19

Ok. So I don't need to prove that it doesn't exist below that? I can just state that?

Answer 20

you could state the function is defined only in the interval x>=-3/2
domain of f
and therefore limit from below doesn't exist

Answer 21

Oh ok. Thanks!

Answer 22

yw!