Find the limits of (x+3)/(x^2-3x-18) = f(x)
lim f(x) if
x->-infinity
x->+infinity
x->-3 (-3 is the hole)
x->+6
x->-6
x->6
Here is the graph :https://www.desmos.com/calculator/cevwm3atol
I am really confused on how to find the limits of an equation with a V.A. and a discontinuity. Any help would be much appreciated!

- anonymous

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- anonymous

It approaches the V.A. which is 6. Is that the limit for x->infinity then?

- anonymous

Unfortunately, I can't quite understand how to find the limit when X is approaching negative infinity due to the fact there is a hole at -3

- Luigi0210

actually no, it is approaching 0

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## More answers

- anonymous

Oh I see, so it is the y value that is the limit? Or am I just on the wrong track completely...

- Luigi0210

When it says "As x is approaching" it means look on the x axis for that value

- Luigi0210

And look for the y value

- terenzreignz

\[\Large \lim_{x\rightarrow \color{red}{\boxed{\color{white}{2}}}}\frac{x+3}{x^2-3x-18}\]

- anonymous

So can I assume that as x approaches negative infinity, it's lim is 0 too?

- Luigi0210

Yup!

- terenzreignz

Was that graph given?

- anonymous

The function is approaching 0 as it approaches negative infinity and as it approaches positive infinity. The function is approaching -.111 as it approaches negative 3. The limit does not exist at 6, because from the right it is approaching positive infinity, and from the left side it is approaching negative infinity. The graph is approaching -.0833 as it approaches -6. The important thing to remember with limits is that it doesn't matter what the function actually equals at the point, it only matters what the function is approaching, just like at -3. There is a vertical asymptote, but that is irrelevant when finding the limit. Type values very close to the x value that you want to approach, and see what the function approaches. Then you will find the limit. For example at positive infinity, type in 100, then 1000, then 10,000 and see that the function is growing continually closer to 0, so the limit is 0.

- anonymous

Woah...

- Luigi0210

I thought you were writing a short story @Parker7e xD

- anonymous

I thought he fell asleep on the keyboard or something...

- anonymous

haha no just a legitimate explanation

- anonymous

the inventor, do you understand?

- cwrw238

note that the expression simplifies to 1/ (x - 6)

- terenzreignz

But if that graph wasn't given... (I think it wasn't) we should know how a function behaves at infinity (or minus infinity).

- terenzreignz

Common trick is to divide each term (in the numerator and denominator) by the largest power of x you see, in this case, 2.
\[\Large \lim_{x\rightarrow \color{red}{\infty}}\frac{x+3}{x^2-3x-18}\cdot \frac{\frac1{x^2}}{\frac1{x^2}}\]\[\Large \lim_{x\rightarrow \color{red}{\infty}}\frac{\frac{1}{x}+\frac3{x^2}}{1-\frac3{x^2}-\frac{18}{x^2}}\]

- terenzreignz

Sorry, slight error with that second bit...
\[\Large \lim_{x\rightarrow \color{red}{\infty}}\frac{\frac{1}{x}+\frac3{x^2}}{1-\frac3{\color{red}x}-\frac{18}{x^2}}\]

- Luigi0210

@terenzreignz I don't think he's that far into calc yet :P

- terenzreignz

Let's not assume... this IS still limits (not derivatives) so what the heck? :P

- anonymous

Parker, why is it approaching -.0833 instead of, say, infinity?
And yes, it is the beginning of pre-calc. We are on our second week ;)

- anonymous

Also, can I award medals to more than one answerer?

- Luigi0210

Sadly, no.

- terenzreignz

But anyway, to wrap it up, THESE terms:
\[\Large \lim_{x\rightarrow \color{red}{\infty}}\frac{\cancel{\frac{1}{x}}^0+\cancel{\frac3{x^2}}^0}{1-\cancel{\frac3{\color{red}x}}^0-\cancel{\frac{18}{x^2}}^0}\]
all go to zero (become really really small, in fact, smaller than any number you could possibly think of) as x goes to infinity.

- terenzreignz

Leaving you with \[\Large \frac01\] which is...
yeah, that :P

- Luigi0210

I think the graph was easier xD

- terenzreignz

But it isn't always going to be available.
Keep that in mind :3

- Luigi0210

And i never learned that method D:

- anonymous

Um, well I think that I am done with math for my lifetime. Maybe Mcdonalds is my best bet...
But as in my previous response, when x->-6, why is it approaching -.0833 instead of, say, infinity?

- terenzreignz

Don't worry... there are more efficient methods anyway... like L'Hopital.
But that's for later. ^_^

- anonymous

I can explain this

- anonymous

Yup, def. going to submit my resume for mcdonalds tonight...

- anonymous

So it is about what the function is approaching at an infinitely close point to the value

- Luigi0210

Come on, calc will get easier, don't give up just yet :)

- anonymous

So if you put -6.00000001 into the function and -6.00000000001 into the function, you will see what the function is very close to, but not yet giving as an output. This is what the limit is. It is what the function is approaching from an infinitely close input. So if you are approaching A, put values close to A in. Say if A is 1, put in .99, then .999, then .9999 into the function to see what the function is getting closer to. Then do it from the other side with 1.1, 1.01, 1.001. This will let you see the behavior of the graph, and what the function is truly APPROACHING. That is the value of the limit.

- anonymous

So here are my answers, correct me if im wrong:
lim f(x) if
x->-infinity is 0
x->+infinity is 0
x->-3 (-3 is the hole) is ...
x->+6 is -infinity
x->-6 +infinity
x->6 is DNE

- terenzreignz

First off, it's not +6 and -6, it's \(\large 6^+\) and \(\large 6^-\) they mean something entirely different :)
Want me to explain?

- Luigi0210

Oh so much explaining :3

- terenzreignz

Don't worry, just listen ^_^

- anonymous

No, I think that I understood that, I just typed it wrong
They indicate the direction right???

- anonymous

Or is mcdonalds in my future?

- terenzreignz

Indeed. What THIS: \(\large 6^+\) means is that x approaches 6 FROM THE RIGHT.

- terenzreignz

Err... that 'indeed' was in response to your previous query (about direction) and not pertaining your future being McDonalds...

- anonymous

What, you DON'T believe I will do good at McDonalds?
Thanks for the optimism...
But is everything else right? I left out x->-3 due to my lack of understanding...

- terenzreignz

Now, take a look at your graph... when the value of x approaches 6 *from the right*, does the function go upward indefinitely or downward indefinitely?

- anonymous

Oh upwards indefinitely...

- terenzreignz

If all goes well and I have my way, you'll be wondering whether you really had a hard time at all ^_^
And yes, upwards. So, the function goes to...? (positive or negative infinity?)

- anonymous

Positive infinity obviously...
Haha this sounds like a teacher explaining to an impossible student
Wait...

- terenzreignz

No... it's a student explaining to another student because the first student knows how hard limits could be to grasp when you're force-fed them early-on and in huge doses...
LOL
okay, what about from the left?

- Luigi0210

^

- anonymous

it goes downwards indefinitely, indicating it is going towards negative infinity, right?
Or does the hole have an effect on this?

- terenzreignz

No. It doesn't :P
Okay, now you know those, care to revise your tentative answer-set? :P

- anonymous

Yes master ;)
x->infinity+ is 0
x->infinity+ is 0
x->-3 (-3 is the hole) is still pretty much unknown to me....
x->6+ is infinity
x->6- is -infinity
x->6 is DNE

- anonymous

I revised the symbols as well

- terenzreignz

Before we deal with that hole... I've been stalki....I mean, looking over the questions you've answered....

- terenzreignz

It seems you rely too much on technology like graphing apps and calculators... we need to fix that.... bit by bit...

- anonymous

Oh that was months ago...
And my teachers promote graphing calculators. I assume you dont? :)

- Luigi0210

Try to avoid them :P

- terenzreignz

I promote awesomeness :3
LOL
anyway
Let's have a look at this.
\[\Large \lim_{x\rightarrow \color{red}{\boxed{\color{white}{2}}}}\frac{x+3}{x^2-3x-18}\]

- terenzreignz

The denominator is factorable, did you notice?

- anonymous

Yes, it becomes -3 and 6

- anonymous

Which I may or may not have just graphed...(Dont worry, I actually know how to factor...)

- terenzreignz

We don't need its roots, we just need its factored form :P
\[\Large \lim_{x\rightarrow \color{red}{\boxed{\color{white}{2}}}}\frac{x+3}{(x+3)(x-6)}\]notice anything?

- anonymous

Ah, I remember that it indicates a hole right?
Also it is now

- anonymous

Wow I am bad at writing equations....
(x+3)/(x+6)

- terenzreignz

Well that's not fair, you cancelled out the x+3 in the denominator but not in the numerator...
I'll see you at the people's court, this is INJUSTICE

- terenzreignz

And besides, it's x-6 in the denominator.
Try again :P

- anonymous

Ah...oh....hmmm I have no idea how I missed that..
Here I have been practicing: "Hello may I take your order?"

- terenzreignz

Yes, I'll have all the correct answers and I want it snappy.

- terenzreignz

Anyway, so it's basically 1/(x-6) yes?

- anonymous

As always, you're right...
Go on...

- terenzreignz

"Calculus" is actually just a pebble (literally, actually) in the bigger mathematical field called "Analysis"
So... let's analyze.

- terenzreignz

You know that when a denominator gets really really small, the whole fraction itself gets bigger and bigger, right?

- anonymous

Yes...

- anonymous

Oh noes... A long reply...

- terenzreignz

Observe...
\[\Large \frac1{\frac12}=2\]\[\Large \frac1{1/1000}=1000\]\[\Large \frac{1}{0.000001}=1000000\]etc

- anonymous

No...stop!
too many 0's!

- terenzreignz

So actually, when a denominator goes to zero, the whole fraction tends to infinity, or gets infinitely large.

- terenzreignz

What happens here
\[\Large \frac1{x-6}\]as x goes to 6 is that the fraction tends to infinity, just don't know WHICH infinity.

- terenzreignz

From the left, x is always less than 6, so x-6 is negative, aye?

- anonymous

Aye, wow...
That makes everything clearer than it should be possible...

- terenzreignz

And, from the right, x is always greater than 6, so x-6 is positive.

- anonymous

Aye

- terenzreignz

Brilliant. So now, it DOES approach infinity as x goes to 6, but from the left, it's always negative, since x-6 is negative and from the right, it's always positive, since x-6 is positive.

- terenzreignz

So, does that clear up why the limit simply doesn't exist as x goes to 6 (as in, you don't specify a direction)

- anonymous

Holy [Insert unholy thing here]! You somehow made that no confusing!

- terenzreignz

no confusing? as in 'not confusing'? You're confusing :/

- anonymous

Real? No, YOU confusing!
;)
Thanks for all your help random stranger on the internet!

- terenzreignz

Terence.
Or TJ.
Anyway, ready to deal with that hole?

- anonymous

Yup!
Now lets see how you can destroy my mind here...

- terenzreignz

Definitely not my intention...Now, obviously, just plugging in x = -3
\[\Large\frac{x+3}{x^2-3x-18}\]
doesn't work here since you get 0/0 right?

- anonymous

Yup, basically undefined

- terenzreignz

No... indeterminate.
0/0, infinity/infinity, 0 times infinity, among others are what are called indeterminate forms, meaning the limit may or may not exist.

- terenzreignz

In this case it does... remember factoring out the denominator?
Do it now.

- anonymous

Pardon, but may I ask why we are doing that again?

- anonymous

uh oh, long reply again...

- terenzreignz

Glad you asked :)
It'd be easier to see what thelimit approaches if you could cancel out that which makes the numerator and denominator equal zero.
After all, if the denominator is zero at x = -3, it can only mean that (x+3) is a factor.

- terenzreignz

I'm going to trust that you'd have no trouble factoring, okay?
\[\Large\frac{x+3}{(x+3)(x-6)}\]

- anonymous

Haha I was trying to figure out how to make a numerator and denominator work in the equation applet

- anonymous

But anyway, go on

- terenzreignz

You and your apps .. -.-
Focus here... something to simplify in
\[\Large\frac{x+3}{(x+3)(x-6)}\]
In particular, something to cancel....

- terenzreignz

We already did this earlier, what's your deal, short-term memory loss? :D

- anonymous

Haha no, applet as in the button you press at the bottom of the reply box...
Sorry, my programming lessons are getting to me

- anonymous

1/(x+6)

- anonymous

pardon, -*

- terenzreignz

\[\Large \frac1{x-6}\]Well then, isn't this better?
Try plugging in x=-3 now and see if it's still indeterminate or undefined?

- anonymous

Ah, 1/-9

- terenzreignz

And that... master inv3ntor, is the limit of \[\Large\frac{x+3}{x^2-3x-18}\] as x goes to minus 3.
Good show, sir ^_^

- anonymous

The funny part is I am british so that actually works pretty well...
And as the stereotype goes: Jolly good show my dear sir! Good show indeed!
Thanks for all the help!

- anonymous

Maybe I WONT work at Mcdonalds!
Wait who am I kidding, I still have to get through the other subjects ;)

- terenzreignz

Well then, you're on your own... for now
I... still have to go to school :P
Cheers, mate ^_^
Signing off
--------------------------------
Terence out

- anonymous

If only every answerer was this in-depth
Adios!

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