anonymous
  • anonymous
FIND the limit!!
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1343661876978:dw|
anonymous
  • anonymous
Ah! This is one best though of visually, let me make you a graph:
anonymous
  • anonymous
ok

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ParthKohli
  • ParthKohli
If you find the function continuous, then you may just plug 3 for \(x\) by the way.
ParthKohli
  • ParthKohli
Let's see how the graph is..
anonymous
  • anonymous
Ok. If I plug in 3, I get 3 sqrt(0)
ParthKohli
  • ParthKohli
Don't do that now.
anonymous
  • anonymous
What you need are numbers slightly less than 3, like 2.9999999999999
ParthKohli
  • ParthKohli
Looks like we can plug in values less than \(3\).
anonymous
  • anonymous
Ok
ParthKohli
  • ParthKohli
Because it is a left handed limit, we may plug numbers like agentx5 said :)
anonymous
  • 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
  • anonymous
ok! can you show how I would solve it
anonymous
  • anonymous
Ok i understand that
anonymous
  • anonymous
About your graph, notice what happens to the function for x > 3
anonymous
  • anonymous
So how would I solve the limit?
ParthKohli
  • ParthKohli
By plugging numbers slightly less than 3.
anonymous
  • anonymous
@telliott99 , I know... look at the notation for imaginary and real...
anonymous
  • anonymous
Just saying the graph is not correct. It doesn't have a hole in it. And do what @ParthKohli says
anonymous
  • anonymous
2.99(9-2.99^2)
ParthKohli
  • 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
  • anonymous
for 2.5, f(x) would be 4.145
ParthKohli
  • ParthKohli
Find them for all.
anonymous
  • 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
  • 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
  • ParthKohli
Calculator, Schmidt.
anonymous
  • anonymous
so agentx5, I should just use 2.999999 to find the limit?
anonymous
  • anonymous
Or if you prefer, 2.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999... Get the point? :P
anonymous
  • anonymous
I got 7.3484
ParthKohli
  • ParthKohli
Just not 3.
ParthKohli
  • ParthKohli
Looks like its getting closer to 7.5, but let's see it for more values.
anonymous
  • anonymous
Ok! I think its going to be 7.5
ParthKohli
  • ParthKohli
Are you sure? Try for 2.9999999 and 2.999999999999999. So on.
anonymous
  • anonymous
I just tried for that
anonymous
  • anonymous
can someone tell me if im right
ParthKohli
  • ParthKohli
I am too lazy to do so :/
anonymous
  • anonymous
come onnn lol
anonymous
  • anonymous
plz! i really need your help..
ParthKohli
  • ParthKohli
All right :) let me check it out on Wolfram.
anonymous
  • anonymous
OK
ParthKohli
  • ParthKohli
@agentx5 Where is the graph?
anonymous
  • 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
  • anonymous
ok thx @agentx5 so what would the limit be for this probb?
ParthKohli
  • ParthKohli
I was actually surprised how I was plugging values and getting it approaching something else. Hence, I asked.
ParthKohli
  • ParthKohli
The limit was getting smaller and smaller. ;)
ParthKohli
  • ParthKohli
You may find from the graph, actually.|dw:1343663313029:dw|
ParthKohli
  • ParthKohli
What is \(y\) approaching in the graph? ^
anonymous
  • anonymous
With the limit shown... |dw:1343663226621:dw|
anonymous
  • anonymous
like 2 or 3 right?
anonymous
  • 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
  • anonymous
Guys do you know the actual limit?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
And I'm trying to help you learn visually what's going on so you're not lost here
anonymous
  • anonymous
Yeah I do see it visually,
anonymous
  • anonymous
is the limit around 2 or 3
anonymous
  • anonymous
wait it looks around 0
anonymous
  • 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
  • anonymous
as 3 approaches 3 from the left, the y value is 0, right?
anonymous
  • anonymous
sorry i meant, as x approaches 3
anonymous
  • 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
  • anonymous
ok so the limit for this function is 0? :D
anonymous
  • 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
  • anonymous
But, limit is 0 right? and yes the ddenominator bneomes 0.. not possible
anonymous
  • 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
  • anonymous
wait so the limit is DNE? or zero!?!
anonymous
  • anonymous
For your original problem, Zero. For the one I just asked you, trying to get you to learn, Does Not Exist.
anonymous
  • anonymous
You will see DNE problems, I can guarantee you.
anonymous
  • anonymous
Oh yeah, I know!:)

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