anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
It approaches the V.A. which is 6. Is that the limit for x->infinity then?
anonymous
  • 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
  • Luigi0210
actually no, it is approaching 0

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anonymous
  • anonymous
Oh I see, so it is the y value that is the limit? Or am I just on the wrong track completely...
Luigi0210
  • Luigi0210
When it says "As x is approaching" it means look on the x axis for that value
Luigi0210
  • Luigi0210
And look for the y value
terenzreignz
  • terenzreignz
\[\Large \lim_{x\rightarrow \color{red}{\boxed{\color{white}{2}}}}\frac{x+3}{x^2-3x-18}\]
anonymous
  • anonymous
So can I assume that as x approaches negative infinity, it's lim is 0 too?
Luigi0210
  • Luigi0210
Yup!
terenzreignz
  • terenzreignz
Was that graph given?
anonymous
  • 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
  • anonymous
Woah...
Luigi0210
  • Luigi0210
I thought you were writing a short story @Parker7e xD
anonymous
  • anonymous
I thought he fell asleep on the keyboard or something...
anonymous
  • anonymous
haha no just a legitimate explanation
anonymous
  • anonymous
the inventor, do you understand?
cwrw238
  • cwrw238
note that the expression simplifies to 1/ (x - 6)
terenzreignz
  • 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
  • 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
  • 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
  • Luigi0210
@terenzreignz I don't think he's that far into calc yet :P
terenzreignz
  • terenzreignz
Let's not assume... this IS still limits (not derivatives) so what the heck? :P
anonymous
  • 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
  • anonymous
Also, can I award medals to more than one answerer?
Luigi0210
  • Luigi0210
Sadly, no.
terenzreignz
  • 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
  • terenzreignz
Leaving you with \[\Large \frac01\] which is... yeah, that :P
Luigi0210
  • Luigi0210
I think the graph was easier xD
terenzreignz
  • terenzreignz
But it isn't always going to be available. Keep that in mind :3
Luigi0210
  • Luigi0210
And i never learned that method D:
anonymous
  • 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
  • terenzreignz
Don't worry... there are more efficient methods anyway... like L'Hopital. But that's for later. ^_^
anonymous
  • anonymous
I can explain this
anonymous
  • anonymous
Yup, def. going to submit my resume for mcdonalds tonight...
anonymous
  • anonymous
So it is about what the function is approaching at an infinitely close point to the value
Luigi0210
  • Luigi0210
Come on, calc will get easier, don't give up just yet :)
anonymous
  • 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
  • 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
  • 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
  • Luigi0210
Oh so much explaining :3
terenzreignz
  • terenzreignz
Don't worry, just listen ^_^
anonymous
  • anonymous
No, I think that I understood that, I just typed it wrong They indicate the direction right???
anonymous
  • anonymous
Or is mcdonalds in my future?
terenzreignz
  • terenzreignz
Indeed. What THIS: \(\large 6^+\) means is that x approaches 6 FROM THE RIGHT.
terenzreignz
  • terenzreignz
Err... that 'indeed' was in response to your previous query (about direction) and not pertaining your future being McDonalds...
anonymous
  • 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
  • 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
  • anonymous
Oh upwards indefinitely...
terenzreignz
  • 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
  • anonymous
Positive infinity obviously... Haha this sounds like a teacher explaining to an impossible student Wait...
terenzreignz
  • 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
  • Luigi0210
^
anonymous
  • anonymous
it goes downwards indefinitely, indicating it is going towards negative infinity, right? Or does the hole have an effect on this?
terenzreignz
  • terenzreignz
No. It doesn't :P Okay, now you know those, care to revise your tentative answer-set? :P
anonymous
  • 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
  • anonymous
I revised the symbols as well
terenzreignz
  • terenzreignz
Before we deal with that hole... I've been stalki....I mean, looking over the questions you've answered....
terenzreignz
  • terenzreignz
It seems you rely too much on technology like graphing apps and calculators... we need to fix that.... bit by bit...
anonymous
  • anonymous
Oh that was months ago... And my teachers promote graphing calculators. I assume you dont? :)
Luigi0210
  • Luigi0210
Try to avoid them :P
terenzreignz
  • 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
  • terenzreignz
The denominator is factorable, did you notice?
anonymous
  • anonymous
Yes, it becomes -3 and 6
anonymous
  • anonymous
Which I may or may not have just graphed...(Dont worry, I actually know how to factor...)
terenzreignz
  • 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
  • anonymous
Ah, I remember that it indicates a hole right? Also it is now
anonymous
  • anonymous
Wow I am bad at writing equations.... (x+3)/(x+6)
terenzreignz
  • 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
  • terenzreignz
And besides, it's x-6 in the denominator. Try again :P
anonymous
  • anonymous
Ah...oh....hmmm I have no idea how I missed that.. Here I have been practicing: "Hello may I take your order?"
terenzreignz
  • terenzreignz
Yes, I'll have all the correct answers and I want it snappy.
terenzreignz
  • terenzreignz
Anyway, so it's basically 1/(x-6) yes?
anonymous
  • anonymous
As always, you're right... Go on...
terenzreignz
  • terenzreignz
"Calculus" is actually just a pebble (literally, actually) in the bigger mathematical field called "Analysis" So... let's analyze.
terenzreignz
  • terenzreignz
You know that when a denominator gets really really small, the whole fraction itself gets bigger and bigger, right?
anonymous
  • anonymous
Yes...
anonymous
  • anonymous
Oh noes... A long reply...
terenzreignz
  • terenzreignz
Observe... \[\Large \frac1{\frac12}=2\]\[\Large \frac1{1/1000}=1000\]\[\Large \frac{1}{0.000001}=1000000\]etc
anonymous
  • anonymous
No...stop! too many 0's!
terenzreignz
  • terenzreignz
So actually, when a denominator goes to zero, the whole fraction tends to infinity, or gets infinitely large.
terenzreignz
  • 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
  • terenzreignz
From the left, x is always less than 6, so x-6 is negative, aye?
anonymous
  • anonymous
Aye, wow... That makes everything clearer than it should be possible...
terenzreignz
  • terenzreignz
And, from the right, x is always greater than 6, so x-6 is positive.
anonymous
  • anonymous
Aye
terenzreignz
  • 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
  • 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
  • anonymous
Holy [Insert unholy thing here]! You somehow made that no confusing!
terenzreignz
  • terenzreignz
no confusing? as in 'not confusing'? You're confusing :/
anonymous
  • anonymous
Real? No, YOU confusing! ;) Thanks for all your help random stranger on the internet!
terenzreignz
  • terenzreignz
Terence. Or TJ. Anyway, ready to deal with that hole?
anonymous
  • anonymous
Yup! Now lets see how you can destroy my mind here...
terenzreignz
  • 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
  • anonymous
Yup, basically undefined
terenzreignz
  • 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
  • terenzreignz
In this case it does... remember factoring out the denominator? Do it now.
anonymous
  • anonymous
Pardon, but may I ask why we are doing that again?
anonymous
  • anonymous
uh oh, long reply again...
terenzreignz
  • 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
  • terenzreignz
I'm going to trust that you'd have no trouble factoring, okay? \[\Large\frac{x+3}{(x+3)(x-6)}\]
anonymous
  • anonymous
Haha I was trying to figure out how to make a numerator and denominator work in the equation applet
anonymous
  • anonymous
But anyway, go on
terenzreignz
  • 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
  • terenzreignz
We already did this earlier, what's your deal, short-term memory loss? :D
anonymous
  • 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
  • anonymous
1/(x+6)
anonymous
  • anonymous
pardon, -*
terenzreignz
  • 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
  • anonymous
Ah, 1/-9
terenzreignz
  • 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
  • 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
  • anonymous
Maybe I WONT work at Mcdonalds! Wait who am I kidding, I still have to get through the other subjects ;)
terenzreignz
  • 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
  • anonymous
If only every answerer was this in-depth Adios!

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