Question about the limit of a square root

[wiz0r]

Hello!

I got the following exercise:

$$\frac{lim}{x\rightarrow6-}$$ $${\sqrt{3x-18}}$$

Now, since I need to evaluate the limit from the function coming from the left to the right that means that I can evaluate the function using x as a value very close to 6, but not six, right? So, since I would have a negative square root to evaluate then the limit does not exist, right?

But, how come when I evaluate that limit on my TI-89 calculator, and some online calculator I get that the limit is zero?

Can someone please explain to me what's going on? I'd be very appreciated. Thank you.

Edwin

 

[rock.freak667]

taking the limit from the left doesn't mean you are to evaluate a negative square root. As you can go from 1,2,3,4,5,6 and as you take values of f(1),f(2),f(3) and so on you will see the limit is zero. [f(x)=√(3x-18)]

 

[wiz0r]

I'm sorry, but I don't really understand.

You see I got an example in my notebook that the professor did, and it is:

$$\frac{lim}{x\rightarrow3-}$$ $${\sqrt{2x-6}}$$

And she told me that the limit didn't exist, but I didn't really understand her explanation. That exercise seems practically the same as the other one, and yet on both exercises the calculator tells me that the limit coming from the left is 0.

I'm really confused, does this means that my professor is wrong? I mean if I evaluate the first exercises in 1, 2, 3, 4, or 5 I would have negative roots, no? Doesn't that tells me that the limit doesn't exists?

 

[Dick]

I'm sorry, but I don't really understand.

You see I got an example in my notebook that the professor did, and it is:

$$\frac{lim}{x\rightarrow3-}$$ $${\sqrt{2x-6}}$$

And she told me that the limit didn't exist, but I didn't really understand her explanation. That exercise seems practically the same as the other one, and yet on both exercises the calculator tells me that the limit coming from the left is 0.

I'm really confused, does this means that my professor is wrong? I mean if I evaluate the first exercises in 1, 2, 3, 4, or 5 I would have negative roots, no? Doesn't that tells me that the limit doesn't exists?

You are right. Your instructor is right. The calculator is wrong. There is no limit from the left. Unless you allow complex roots. But I don't think that's what the question is about.

 

[rock.freak667]

I'm sorry, but I don't really understand.

You see I got an example in my notebook that the professor did, and it is:

$$\frac{lim}{x\rightarrow3-}$$ $${\sqrt{2x-6}}$$

And she told me that the limit didn't exist, but I didn't really understand her explanation. That exercise seems practically the same as the other one, and yet on both exercises the calculator tells me that the limit coming from the left is 0.

I'm really confused, does this means that my professor is wrong? I mean if I evaluate the first exercises in 1, 2, 3, 4, or 5 I would have negative roots, no? Doesn't that tells me that the limit doesn't exists?

Forget what I said about the negative square root thing.

For lim(x→a) f(x) to exist then lim(x →a⁺)f(x) = lim(x→a^-)f(x)

This basically means that for the limit to exist, the limit as you approach 'a' from the left must be the as if you approached 'a' from the right.

Now in your example: lim(x→6^-)√(3x-18), if we use complex numbers (i=√-1),

if we compute from 1,2,3,4,5,6

we can see that

1: √(-15) =√15i

2: √(-12)=√12i

3: √(-9)=3i

4: √(-6)=√6i

5: √(-3)=√3i

6: √0 = 0

so you can see as x goes from 1 to 6, the complex numbers are getting lesser and lesser and are tending to zero. So you can say that the limit as x→6^- of √(3x-18) is 0


EDIT: Dick is right if you are only meant to use real numbers. I used complex numbers, if you didn't do this topic yet, then disregard my post.