FIND the limit!!

- anonymous

FIND the limit!!

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- anonymous

|dw:1343661876978:dw|

- anonymous

Ah! This is one best though of visually, let me make you a graph:

- anonymous

ok

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## More answers

- ParthKohli

If you find the function continuous, then you may just plug 3 for \(x\) by the way.

- ParthKohli

Let's see how the graph is..

- anonymous

Ok. If I plug in 3, I get 3 sqrt(0)

- ParthKohli

Don't do that now.

- anonymous

What you need are numbers slightly less than 3, like 2.9999999999999

- ParthKohli

Looks like we can plug in values less than \(3\).

- anonymous

Ok

- ParthKohli

Because it is a left handed limit, we may plug numbers like agentx5 said :)

- anonymous

That - sign above the limit means "numbers from the left hand side limit" or the number slightly less than 3 (but as close as you can get)

- anonymous

ok! can you show how I would solve it

- anonymous

Ok i understand that

- anonymous

About your graph, notice what happens to the function for x > 3

- anonymous

So how would I solve the limit?

- ParthKohli

By plugging numbers slightly less than 3.

- anonymous

@telliott99 , I know... look at the notation for imaginary and real...

- anonymous

Just saying the graph is not correct. It doesn't have a hole in it. And do what @ParthKohli says

- anonymous

2.99(9-2.99^2)

- ParthKohli

\[ \begin{array}{l|r} x & f(x) \\ \hline 2.5 & \\ 2.9 \\ 2.999 \\ 2.99999\end{array} \] Fill the f(x)'s.

- anonymous

for 2.5, f(x) would be 4.145

- ParthKohli

Find them for all.

- anonymous

And and as you get closer and closer to 2.999999999999? Every closer to 3 from the "just under" side and you'll approach some limit.

- anonymous

Ok can you help find the limit now? Like I understand how to find that, i just dont haeve time to find them all

- ParthKohli

Calculator, Schmidt.

- anonymous

so agentx5, I should just use 2.999999 to find the limit?

- anonymous

Or if you prefer, 2.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...
Get the point? :P

- anonymous

I got 7.3484

- ParthKohli

Just not 3.

- ParthKohli

Looks like its getting closer to 7.5, but let's see it for more values.

- anonymous

Ok! I think its going to be 7.5

- ParthKohli

Are you sure? Try for 2.9999999 and 2.999999999999999.
So on.

- anonymous

I just tried for that

- anonymous

can someone tell me if im right

- ParthKohli

I am too lazy to do so :/

- anonymous

come onnn lol

- anonymous

plz! i really need your help..

- ParthKohli

All right :) let me check it out on Wolfram.

- anonymous

OK

- ParthKohli

@agentx5 Where is the graph?

- anonymous

Here's the algebra trick, you need to find what y would be when x=3. You can't do this normally but...
\(x\sqrt{(3+x)(3-x)}\) , For what solutions does y = 0? Well, 0, +3 and -3. So... ;-)
Replacing with a better graph (more accurate, real #'s only):
|dw:1343662913297:dw|

- anonymous

ok thx @agentx5 so what would the limit be for this probb?

- ParthKohli

I was actually surprised how I was plugging values and getting it approaching something else. Hence, I asked.

- ParthKohli

The limit was getting smaller and smaller. ;)

- ParthKohli

You may find from the graph, actually.|dw:1343663313029:dw|

- ParthKohli

What is \(y\) approaching in the graph? ^

- anonymous

With the limit shown...
|dw:1343663226621:dw|

- anonymous

like 2 or 3 right?

- anonymous

I actually had my imaginary lines going the wrong way @telliott99 and the circles were too high in the original sketch (they're on the x-axis), I stand corrected. Although I'm not sure that's what you meant. ;-)

- anonymous

Guys do you know the actual limit?

- anonymous

Yes.

- anonymous

And I'm trying to help you learn visually what's going on so you're not lost here

- anonymous

Yeah I do see it visually,

- anonymous

is the limit around 2 or 3

- anonymous

wait it looks around 0

- anonymous

\[\lim_{x \rightarrow 3^-}\] What is the output of the function approaching? The input gets closer to 3 from the "slightly less" side.

- anonymous

as 3 approaches 3 from the left, the y value is 0, right?

- anonymous

sorry i meant, as x approaches 3

- anonymous

Or using the analytical method, as you plug in larger and larger 2.9999999999999999...'s you get smaller and smaller numbers. Infinitesimally small. To the point gets damned close to zero. Thus the limit is zero, correct.

- anonymous

ok so the limit for this function is 0? :D

- anonymous

The reason you can't just substitute 3 in place of x, officially is because you might have one like this:
\[\lim_{x \rightarrow 1} = \frac{1}{1-x}\]
Look what happens if you plug in 1?

- anonymous

But, limit is 0 right? and yes the ddenominator bneomes 0.. not possible

- anonymous

In this case your left and your right hand limits are going in opposite directions, and the limit doesn't exist.
|dw:1343663845294:dw|
And yes, twice to the above post. :-)

- anonymous

wait so the limit is DNE? or zero!?!

- anonymous

For your original problem, Zero.
For the one I just asked you, trying to get you to learn, Does Not Exist.

- anonymous

You will see DNE problems, I can guarantee you.

- anonymous

Oh yeah, I know!:)

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