Question

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!

Answer 1

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

Answer 2

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

Answer 3

actually no, it is approaching 0

Answer 4

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

Answer 5

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

Answer 6

And look for the y value

Answer 7

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

Answer 8

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

Answer 9

Yup!

Answer 10

Was that graph given?

Answer 11

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.

Answer 12

Woah...

Answer 13

I thought you were writing a short story @Parker7e xD

Answer 14

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

Answer 15

haha no just a legitimate explanation

Answer 16

the inventor, do you understand?

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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 ;)

Answer 24

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

Answer 25

Sadly, no.

Answer 26

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.

Answer 27

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

Answer 28

I think the graph was easier xD

Answer 29

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

Answer 30

And i never learned that method D:

Answer 31

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?

Answer 32

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

Answer 33

I can explain this

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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.

Answer 38

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

Answer 39

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

Want me to explain?

Answer 40

Oh so much explaining :3

Answer 41

Don't worry, just listen ^_^

Answer 42

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

Answer 43

Or is mcdonalds in my future?

Answer 44

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

Answer 45

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

Answer 46

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...

Answer 47

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?

Answer 48

Oh upwards indefinitely...

Answer 49

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?)

Answer 50

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

Answer 51

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...
<bitter>

LOL
okay, what about from the left?

Answer 52

^

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

I revised the symbols as well

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

Try to avoid them :P

Answer 61

I promote awesomeness :3

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

Answer 62

The denominator is factorable, did you notice?

Answer 63

Yes, it becomes -3 and 6

Answer 64

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

Answer 65

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?

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

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

Answer 72

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

Answer 73

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

Answer 74

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

Answer 75

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

Answer 76

Yes...

Answer 77

Oh noes... A long reply...

Answer 78

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

Answer 79

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

Answer 80

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

Answer 81

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.

Answer 82

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

Answer 83

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

Answer 84

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

Answer 85

Aye

Answer 86

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.

Answer 87

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)

Answer 88

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

Answer 89

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

Answer 90

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

Answer 91

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

Answer 92

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

Answer 93

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?

Answer 94

Yup, basically undefined

Answer 95

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.

Answer 96

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

Answer 97

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

Answer 98

uh oh, long reply again...