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FIND the limit!!

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

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Other answers:

If you find the function continuous, then you may just plug 3 for \(x\) by the way.
Let's see how the graph is..
Ok. If I plug in 3, I get 3 sqrt(0)
Don't do that now.
What you need are numbers slightly less than 3, like 2.9999999999999
Looks like we can plug in values less than \(3\).
Because it is a left handed limit, we may plug numbers like agentx5 said :)
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)
ok! can you show how I would solve it
Ok i understand that
About your graph, notice what happens to the function for x > 3
So how would I solve the limit?
By plugging numbers slightly less than 3.
@telliott99 , I know... look at the notation for imaginary and real...
Just saying the graph is not correct. It doesn't have a hole in it. And do what @ParthKohli says
\[ \begin{array}{l|r} x & f(x) \\ \hline 2.5 & \\ 2.9 \\ 2.999 \\ 2.99999\end{array} \] Fill the f(x)'s.
for 2.5, f(x) would be 4.145
Find them for all.
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.
Ok can you help find the limit now? Like I understand how to find that, i just dont haeve time to find them all
Calculator, Schmidt.
so agentx5, I should just use 2.999999 to find the limit?
Or if you prefer, 2.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999... Get the point? :P
I got 7.3484
Just not 3.
Looks like its getting closer to 7.5, but let's see it for more values.
Ok! I think its going to be 7.5
Are you sure? Try for 2.9999999 and 2.999999999999999. So on.
I just tried for that
can someone tell me if im right
I am too lazy to do so :/
come onnn lol
plz! i really need your help..
All right :) let me check it out on Wolfram.
@agentx5 Where is the graph?
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|
ok thx @agentx5 so what would the limit be for this probb?
I was actually surprised how I was plugging values and getting it approaching something else. Hence, I asked.
The limit was getting smaller and smaller. ;)
You may find from the graph, actually.|dw:1343663313029:dw|
What is \(y\) approaching in the graph? ^
With the limit shown... |dw:1343663226621:dw|
like 2 or 3 right?
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. ;-)
Guys do you know the actual limit?
And I'm trying to help you learn visually what's going on so you're not lost here
Yeah I do see it visually,
is the limit around 2 or 3
wait it looks around 0
\[\lim_{x \rightarrow 3^-}\] What is the output of the function approaching? The input gets closer to 3 from the "slightly less" side.
as 3 approaches 3 from the left, the y value is 0, right?
sorry i meant, as x approaches 3
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.
ok so the limit for this function is 0? :D
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?
But, limit is 0 right? and yes the ddenominator bneomes 0.. not possible
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. :-)
wait so the limit is DNE? or zero!?!
For your original problem, Zero. For the one I just asked you, trying to get you to learn, Does Not Exist.
You will see DNE problems, I can guarantee you.
Oh yeah, I know!:)

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