/sci/ - Science & Math

[Anonymous No.2938907]

What is the use of a limit? If I say, f(x) = (x^2, x =/= 2) (3, x = 2)
and graph that, it would give me an identical result to lim x -> 2
f(x) = ?

[Anonymous No.2938960]

Well, we need limits to define derivatives. So I'd say they're pretty useful.

[Anonymous No.2938955]

bump

[Anonymous No.2938970]

Limits aren't good for anything. It's just some kind oftrolling.

[Anonymous No.2938991]

>>2938907
yeah, pretty much to definition. We try hard as fuck to make mathematics an exact science, limits are the way to explain the "kinda" type of solutions, like when we use infinity. We can't really divide 5 with infinity, but we can have limits, and that is zero.

[Anonymous No.2938999]

>>2938991
This is a terrible answer. There's nothing "kinda" about them. Limits are fundamental to tons of rigorously developed concepts.

[Anonymous No.2939014]

Limits are used to explain something that doesn't fit into the basic believes of mathematicians. The fundamental problem is that these atheist can't explain it with god.

[Anonymous No.2939032]

Limits are useful when you are dealing with undefined regions of a graph. For instance, a graph of f(x) = <span class="math">e^x[/spoiler]

http://www.wolframalpha.com/input/?i=y+%3D+e^x

Shows you that the graph gets very close to the x-axis but never touches it.

We can say that as the x value 'goes' to negative infinity, the y value is actually proceeding to (but never getting to) 0.

Limits have other uses but for beginners, it's best to look at things like asymptotes: especially division by zero asymptotic graphs.

[Anonymous No.2939040]

2/x
lim as x -> 0 = infinity

[Anonymous No.2939059]

>mrw /sci/ can't explain limits

[Anonymous No.2939057]

limit as x->1 of f(x)=1
x never actually reaches 1, but it's limit does
x basically =.999... in this case, but never =1

[Anonymous No.2939083]

lim faggotry->infinity = OP

[Anonymous No.2939076]

>my reaction when /sci/ can't explain limits

[Scientist !!ThFjnJh4EkH No.2939097]

>>2939076
I can, and your pic is actually wrong.

lim as x->8 of [ 1 / (x - 8) ] does not go to infinity. It is undefined.

[Anonymous No.2939110]

>>2939032
I find limits very hard to grasp. How do you read it when you see e.g lim x->2 x = 5?

And what is it supposed to answer? In your example, would you read it: "As x approaches negative infinity..." and then what does the e^x mean, and how do you read and use it?

[Anonymous No.2939149]

>>2939097

[Scientist !!ThFjnJh4EkH No.2939157]

>>2939149
What? Want a rigorous proof? lim as x->8 of [1/(x-8)] does not go to infinity. What are you trying to say?

[Anonymous No.2939167]

>>2939157

you fail so hard, i can't even find the fitting face palm picture

[Anonymous No.2939177]

>>2939157
f(x)=1/(x-8) when x=8 is undefined
I think that's what you're thinking of

[Anonymous No.2939230]

>>2939110

You are looking at what the y-values are doing as you plug in x-values.

You see, <span class="math">f(x) = e^x[/spoiler] is a function with respect to x. That means the y-value (f(x) can be written as y too) is dependent on what you are doing to the x-value.

So for <span class="math">\displaystyle y = e^x[/spoiler], this is what happens as you plug in more negative x-values:

x-values
-1 (so you're plugging in -1: <span class="math">y = e^{-1}[/spoiler])
-2
-10
-20

y-values
0.367879
0.135335
0.0000453999
2.06115x10^-9

You can see how close y is getting to zero even negative 20 units out. If you were to plug in something like -2375, the y-value would still not be zero but it would have 50 or so trailing zeros in the decimal place before any other numbers were visible (0.000000000000000000000..00234 or something).

Has your instructor taught you how to solve for limits using the table approach?

This doesn't make much sense:

<span class="math">\displaystyle \lim_{x\to2} x = 5[/spoiler]

It would read as "the limit as x approaches 2 for the function x is 5"

Maybe you meant:

<span class="math">\displaystyle \lim_{x\to2} x[/spoiler]

<span class="math">\displaystyle L = 2[/spoiler]

[Anonymous No.2939246]

>>2939097
>mfw you forgot you learned left and right-sided limits which is probably what this instructor was teaching

[Anonymous No.2939308]

>>2939230
So we're trying to find what f(x) equals, and that is the limit?

[Anonymous No.2939349]

the answer that applies to real life use of math by engineers, scientists, financial mathematics, various unrelated computer code, etc...
because ultimately pretty much everything is represented by a Fourier Transform (not necessarily the specific sine/cosine/complex-exponential, but things like spherical harmonics/associate legendre, bessell, hermite/laguerre, airy functions, etc.)


you represent a function by a truncated infinite series.


limits are tools that are used in this context to ensure that the functional form actually approximates the behavior of the function, as well as certain conditions like convergence (which means that the sum you use does not add up to infinity), existence, etc.


ultimately these concepts are tied into the pure mathematics that derive the results.


functional analysis, analytic functions, etc.


those are just extremely widely used examples of the "tool" that is a limit.


besides, the Limit is a fundamental concept in calculus.

the derivative and the integral are defined in terms of limits.


without the limit, they would not exist.

[Anonymous No.2939360]

Limits allow you to define the behavior of a function as its input becomes arbitrarily close to a certain real number, or as it goes off to infinity.

[Anonymous No.2939372]

>>2939308
Also, could we say that:
"The limit as x approaches 5 for f(x)"
Then L = 5?

[Anonymous No.2939470]

>>2939308

Sort of...try this:

http://tutorial.math.lamar.edu/Classes/CalcI/Tangents_Rates.aspx

That's the entire introduction to the limit process. If you want to skip to the limit process, try this:

http://tutorial.math.lamar.edu/Classes/CalcI/TheLimit.aspx

Look at the first example:

<span class="math">\displaystyle lim_{x\to1} f(x) = \frac{2 - 2x^2}{x - 1}[/spoiler]

Explain how he arrived at the answer of <span class="math">L = -4[/spoiler].

Here's what you could do:

1) Graph the function and see what it's doing.

http://www.wolframalpha.com/input/?i=%282-2x^2%29%2F%28x-1%29

As the x-value is approaching 1, what is the y-value doing?

2) Plug in some values around the limit but not exactly on the limit.

See, if you tried to plug in 1 in this case you'd get undefined in the denominator. So choose values around 1 on both sides.

x values on the left, y values on the right:

0.5 = -3
0.75 = -3.5
0.99 = -3.98
1.25 = -4.5
1.50 = -5
1.01 = -4.02

You can see that the closer x gets to 1, the closer y gets to 4.

3) Evaluate it analytically.

Believe it or not, you can manipulate this to make more sense.

<span class="math">\displaystyle \displaystyle lim_{x\to1} \frac{2 - 2x^2}{x - 1}[/spoiler]

Yes, there's a discrepancy in the denominator. A hole. See if you can remove the hole.

<span class="math">\displaystyle lim_{x\to1} \frac{2 - 2x^2}{x - 1}[/spoiler]
<span class="math">\displaystyle lim_{x\to1} \frac{-2x^2 + 2}{x - 1}[/spoiler]
<span class="math">\displaystyle lim_{x\to1} \frac-2{(x^2 - 2)}{x - 1}[/spoiler]
<span class="math">\displaystyle lim_{x\to1} \frac-2{(x - 1)(x + 1)}{x - 1}[/spoiler]

Well look at that. You can get rid of the hole.

<span class="math">\displaystyle lim_{x\to1} -2(x + 1)[/spoiler]

Evaluate the limit.

<span class="math">\displaystyle lim_{x\to1} -2(x+1)[/spoiler]

Plug in 1.

<span class="math">\displaystyle L = -2(1+1)[/spoiler]
<span class="math">\displaystyle L = -4[/spoiler]

[Anonymous No.2939481]

>>2939308

Sort of...try this:

http://tutorial.math.lamar.edu/Classes/CalcI/Tangents_Rates.aspx

That's the entire introduction to the limit process. If you want to skip to the limit process, try this:

http://tutorial.math.lamar.edu/Classes/CalcI/TheLimit.aspx

Look at the first example:

<span class="math">\displaystyle lim_{x\to1} f(x) = \frac{2 - 2x^2}{x - 1}[/spoiler]

Explain how he arrived at the answer of <span class="math">L = -4[/spoiler].

Here's what you could do:

1) Graph the function and see what it's doing.

http://www.wolframalpha.com/input/?i=%282-2x^2%29%2F%28x-1%29

As the x-value is approaching 1, what is the y-value doing?

2) Plug in some values around the limit but not exactly on the limit.

See, if you tried to plug in 1 in this case you'd get undefined in the denominator. So choose values around 1 on both sides.

x values on the left, y values on the right:

0.5 = -3
0.75 = -3.5
0.99 = -3.98
1.25 = -4.5
1.50 = -5
1.01 = -4.02

You can see that the closer x gets to 1, the closer y gets to 4.

3) Evaluate it analytically.

Believe it or not, you can manipulate this to make more sense.

<span class="math">\displaystyle \displaystyle lim_{x\to1} \frac{2 - 2x^2}{x - 1}[/spoiler]

Yes, there's a discrepancy in the denominator. A hole. See if you can remove the hole.

<span class="math">\displaystyle lim_{x\to1} \frac{2 - 2x^2}{x - 1}[/spoiler]
<span class="math">\displaystyle lim_{x\to1} \frac{-2x^2 + 2}{x - 1}[/spoiler]
<span class="math">\displaystyle lim_{x\to1} -2\frac{(x^2 - 2)}{x - 1}[/spoiler]
<span class="math">\displaystyle lim_{x\to1} -2\frac{(x - 1)(x + 1)}{x - 1}[/spoiler]

Well look at that. You can get rid of the hole.

<span class="math">\displaystyle lim_{x\to1} -2(x + 1)[/spoiler]

Evaluate the limit.

<span class="math">\displaystyle lim_{x\to1} -2(x+1)[/spoiler]

Plug in 1.

<span class="math">\displaystyle L = -2(1+1)[/spoiler]
<span class="math">\displaystyle L = -4[/spoiler]

[Scientist !!ThFjnJh4EkH No.2939489]

First, easy proof:

lim as x -> 8+ of 1/(x-8) = - inf
lim as x -> 8- of 1/(x-8) = inf
Thus the limit is undefined.

Second, longer more formal proof:

Let me remind you of the definition:

lim as x -> C of f(x) goes to infinity, written colloquially as lim as x -> C of f(x) = inf, is true iff
(for all E in R) (there exists a D in R+) (for all x in domain of f) (0 < |x - C| < D implies f(x) > E)

Now, let's begin the proof. It'll be a proof by contradiction.

[continued next post]

[Scientist !!ThFjnJh4EkH No.2939492]

>>2939489
Assume that lim as x->8 of [(1/(8-x)] = inf.
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f) (0 < |x-8| < D implies f(x) > E)
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f) (0 < |x-8| < D implies 1/(x-8) > E)
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) (0 < |x-8| < D implies 1/(x-8) > E)
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) (|x-8| < D implies 1/(x-8) > E)
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) (|x-8| < D implies 1/E <= x-8)
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) (-(x-8) < D implies 1/E <= x-8)
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) (x-8 >= -D implies 1/E <= x-8)
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) not (x-8 >= -D) OR 1/E <= x-8
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) not not [not (x-8 >= -D) OR 1/E <= x-8]
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) not [x-8 >= -D AND 1/E > x-8]
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) not [1/E >= -D]
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) 1/E < -D
Thus (for all E in R) (there exists a D in R+) (for all x in domain of f and less than 8) 1/E < -D

Noting that [domain of f and less than 8] is non-empty, we can get:
(for all E in R) (there exists a D in R+) 1/E < -D

Thus (there exists a D in R+) 1/1 < -D
Thus (there exists a D in R+) -1 > D
Thus there exists a positive Real less than -1
False

Thus an assumption was false.
Thus that lim as x->8 of [(1/(8-x)] does not go infinity.

[Anonymous No.2939519]

>>2939481

I'm sorry this should be -4 not +4:

>You can see that the closer x gets to 1, the closer y gets to -4.

[Anonymous No.2939534]

>>2939489
>>2939492

You shut your whoreface. It was supposed to be a funny picture so I didn't actually evaluate the content for validity.

[Anonymous No.2939564]

>>2939519
Yes, yes... I kind of understand. Kind of. Thanks for taking the time to write all that.

So f(x) is basically the y value we're trying to find?
What about:
"The limit as x approaches 9 for f(x) = 18"? Is that even possible? Are there other online resources to help me understand this?

[Anonymous No.2939638]

>>2939564

Yes you are essentially plugging in the x-values. However, you will come to a problem when you are asked to evaluate limits at infinity.

You cannot 'plug' in infinity since it isn't considered a number. You can only view what the graph is doing and how the x and y values are behaving.

This doesn't make sense:

>"The limit as x approaches 9 for f(x) = 18"? Is that even possible? Are there other online resources to help me understand this?

The limit of that is just 18. You can't plug in any x-value that will change the 18 because...there isn't an x-value. =) You could do this:

<span class="math">\displaystyle lim_{x\to9} f(x) = \frac{18}{x}[/spoiler]
<span class="math">\displaystyle L = 2[/spoiler]

You would just plug in 9 and get 18/9 which is 2 and that's the limit.

But what happens when you get something like this?

<span class="math">\displaystyle lim_{x\to0} f(x) = cos(\frac{\pi}{x})[/spoiler]

Rut-roh. Can't really eyeball that. If you were to use the table approach or graph this, you would see x does not really approach any single value.

If you try to evaluate this analytically, you can't because there is no manipulation you can do to make the hole go away.

There is no limit for <span class="math">\displaystyle lim_{x\to0} f(x) = cos(\frac{\pi}{x})[/spoiler]. It does not exist.

You could be asked to find the limit of this:

<span class="math">\displaystyle lim_{x\to\infty} f(x) = cos(\frac{\pi}{x})[/spoiler]

This is basically asking you what is happening to this function as the denominator blows up. As you begin to pick x values, you will see that you are just dividing <span class="math">\pi[/spoiler] by larger numbers which is making <span class="math">\pi[/spoiler] even smaller.

The cosine of a very, very small positive value is going to be closer to 1 and that is the limit:

<span class="math">\displaystyle lim_{x\to\infty} f(x) = cos(\frac{\pi}{x})[/spoiler]
<span class="math">\displaystyle L = 1[/spoiler]

I don't have any other resources besides

http://www.khanacademy.org/

[Anonymous No.2940263]

hahahahaha, a really simple and honest question that was answered in the first post became a shitstorm.

You guys are retarded, and it isn't because of your grasp of mathematics but because how you react at things in the internet

[Anonymous No.2942470]

>>2940263

What shitstorm? Almost every post in this thread is useful: except yours.

[Anonymous No.2942493]

>>2940263
how the hell is "bump" the answer? you are a retard, also this:>>2942470

[Anonymous No.2942498]

>>2939638
Wait... The limit as X approaches infinity of any constant over X is 0, right? Not 1.

[Anonymous No.2942509]

>>2942498
Me again. Disregard my last post: Cos is indeed 1 as x approaches infinity.

[Anonymous No.2942636]

>>2939638
what happens is that you take the limit of the inside, obviously lim x->inf of pi/x is zero and lim u->0 of cos(u) is 1. :P

[Anonymous No.2942766]

OP here: So we're basically trying to find what value y approaches using the f(x) function? As in;
The limit as x approaches 2 for x
Then L = 2 because the y value is approaching 2 as well?

[Anonymous No.2942802]

>>2942766

Bingo. Just remember that f(x) simply means f as a function of x and is really just y. It's easy to confuse notation if you use f(x) without knowing what it means.

Read this:

http://www.purplemath.com/modules/fcnnot.htm

Variables are more flexible, easier to read, and can give you more information.

The same is true of "y" and "f(x)" (pronounced as "eff-of-eks"). For functions, the two notations mean the exact same thing, but "f(x)" gives you more flexibility and more information. You used to say "y = 2x + 3; solve for y when x = –1". Now you say "f(x) = 2x + 3; find f(–1)" (pronounced as "f-of-x is 2x plus three; find f-of-negative-one"). You do exactly the same thing in either case: you plug in –1 for x, multiply by 2, and then add the 3, simplifying to get a final value of +1.

[Anonymous No.2942860]

>>2942802
So if we would have:
"The limit as x -> -1 for f(x) = 2x + 3"
Then L = 1
?

[Anonymous No.2942868]

>>2942860

http://www.wolframalpha.com/input/?i=The+limit+as+x+-%3E+-1+for+f%28x%29+%3D+2x+%2B+3

[Anonymous No.2942883]

>>2942868
Woohoo!

Anyone else got anything I should read to reinforce this idea, or make it more intuitive?

[Anonymous No.2943010]

Not OP

Basically limits are a way of saying: As I put in these values for x up to a certain number, this is what will happen to the y value.

Is this correct?

[Anonymous No.2943270]

>>2943010
Basically. You're trying to find what value y approaches.