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Can someone please explain limits to me?

I get the jist of it, as a value of x (or any other variable) approaches some number, the function's value will get closer to some other value.

Or is this not correct?

I'm asking as I just want to broaden my knowledge of maths :cool::cool:

(Also, how do you USE limits in problems?)

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Reply 1
Original post by tohaaaa
I get the jist of it, as a value of x (or any other variable) approaches some number, the function's value will get closer to some other value.

Or is this not correct?

I'm asking as I just want to broaden my knowledge of maths :cool::cool:

(Also, how do you USE limits in problems?)


In a nutshell, yeah (though does infinity count as a number? :rolleyes:)
Limits are used in things like graph sketching, and they can be used as part of a problem, i.e. evaluating the limit can be the problem.
Limits are used when you can't evaluate a problem at an exact point, because problems arise.
e,g.
limx1x2x1\displaystyle\lim_{x \to 1} \dfrac{x^2}{x-1}
Reply 2
Original post by joostan
In a nutshell, yeah (though does infinity count as a number? :rolleyes:)
Limits are used in things like graph sketching, and they can be used as part of a problem, i.e. evaluating the limit can be the problem.
Limits are used when you can't evaluate a problem at an exact point, because problems arise.
e,g.
limx1x2x1\displaystyle\lim_{x \to 1} \dfrac{x^2}{x-1}


Ah, so that equation there, x can never be 1 because (1-1 = 0, which means you'd have to divide by 0)... but how do you find limits? (or is that hard?)
Reply 3
Original post by tohaaaa
Ah, so that equation there, x can never be 1 because (1-1 = 0, which means you'd have to divide by 0)... but how do you find limits? (or is that hard?)


limxc\displaystyle\lim_{x \to c}

c can be any number. So there may be easy limits:

limx220x1\displaystyle\lim_{x \to 2} \dfrac{20}{x-1} would simply yield 20
.
If you are wondering about finding certain values of c, look for things that would, outside the world of limits, cause headaches :laugh:
Reply 4
Original post by tohaaaa
Ah, so that equation there, x can never be 1 because (1-1 = 0, which means you'd have to divide by 0)... but how do you find limits? (or is that hard?)

Well, the example I gave is fairly straightforward, it's just infinity or -infinity depending on the direction of approach to 1.
limx1x21x1\displaystyle\lim_{x \to 1}\dfrac{x^2-1}{x-1}
is a different story, a bit of manipulation can yield a limit.

For more complicated limits there's L'Hopital's rule for indeterminate quotients helps to evaluate some.
It states that:
limxcf(x)g(x)=limxcf(x)g(x)\displaystyle\lim_{x \to c} \dfrac{f(x)}{g(x)}=\displaystyle\lim_{x \to c}\dfrac{f'(x)}{g'(x)}
Where c is just a constant.
The limit is also used in differentiation from first principles. :smile:
Reply 5
Original post by joostan
Well, the example I gave is fairly straightforward, it's just infinity or -infinity depending on the direction of approach to 1.
limx1x21x1\displaystyle\lim_{x \to 1}\dfrac{x^2-1}{x-1}
is a different story, a bit of manipulation can yield a limit.

For more complicated limits there's L'Hopital's rule for indeterminate quotients helps to evaluate some.
It states that:
limxcf(x)g(x)=limxcf(x)g(x)\displaystyle\lim_{x \to c} \dfrac{f(x)}{g(x)}=\displaystyle\lim_{x \to c}\dfrac{f'(x)}{g'(x)}
Where c is just a constant.
The limit is also used in differentiation from first principles. :smile:


Woah, you lost me with the second paragraph :smile:

So in short, is a limit a value that a function approaches as one of the variables in that function approaches some value.
Reply 6
Original post by hslakaal
limxc\displaystyle\lim_{x \to c}

c can be any number. So there may be easy limits:

limx220x1\displaystyle\lim_{x \to 2} \dfrac{20}{x-1} would simply yield 20
.
If you are wondering about finding certain values of c, look for things that would, outside the world of limits, cause headaches :laugh:


Ok, so in your second example, as x approaches 2, 20/x-1 will approach 20?
Reply 7
Original post by tohaaaa
Woah, you lost me with the second paragraph :smile:


Ah, OK. Never mind :lol:

So in short, is a limit a value that a function approaches as one of the variables in that function approaches some value.

More or less :smile:
Reply 8
Original post by tohaaaa
Woah, you lost me with the second paragraph :smile:

So in short, is a limit a value that a function approaches as one of the variables in that function approaches some value.


Yeap.

Unless you're doing further maths at a-level or above, you do not need to know L'Hopital's rule
Reply 9
Original post by hslakaal
Yeap.

Unless you're doing further maths at a-level or above, you do not need to know L'Hopital's rule


Well, I plan to do further maths :colondollar::colondollar::colondollar::colondollar::colondollar:

Soo..... how hard is it to understand?
Reply 10
Original post by tohaaaa
Ok, so in your second example, as x approaches 2, 20/x-1 will approach 20?


Yes. It's simply a way of writing that the variable will never reach a certain constant (in this case 2), and hence your "answer" will be near 20, which for all reasons and purposes is identical to 20 since it's a limit.
Reply 11
Original post by tohaaaa
Woah, you lost me with the second paragraph :smile:

So in short, is a limit a value that a function approaches as one of the variables in that function approaches some value.


Yes, exactly. limx1x=1,  limx1x=0\displaystyle \lim_{x \to 1} x = 1,\ \ \lim_{x \to \infty} \dfrac{1}{x} = 0; if we let f(0)=1f(0) = 1 and f(x)=0f(x) = 0 everywhere, then limx0f(x)=0\lim_{x \to 0} f(x) = 0 , because however close we get to 0, f(x) is still 0.
Reply 12
This is the problem with young people these days... they just don't know their limits...
Reply 13
Original post by tohaaaa
Well, I plan to do further maths :colondollar::colondollar::colondollar::colondollar::colondollar:

Soo..... how hard is it to understand?


It isn't very hard.

Though the pitfalls of maths at GCSE and up seems to ignore teaching the proof of differentiation using limits which I had always been puzzled by.

It essentially builds on the definition of differential maths. once you get to it, it'll be easy :smile:
Reply 14
Original post by hslakaal
It isn't very hard.

Though the pitfalls of maths at GCSE and up seems to ignore teaching the proof of differentiation using limits which I had always been puzzled by.

It essentially builds on the definition of differential maths. once you get to it, it'll be easy :smile:


They don't even teach basic differentiation at GCSE, I doubt anyone would understand the proof!
Reply 15
Original post by tohaaaa
They don't even teach basic differentiation at GCSE, I doubt anyone would understand the proof!

The concept of differentiation and why it works is easy to understand. It's probably on the same difficulty as being able to notice that if you increase the number of sides of a polygon to infinity, your shape starts to look like a circle.

The reason we teach concepts of calculus at a higher level than basic arithmetic is because of the computation.
Reply 16
Original post by 0x2a
The concept of differentiation and why it works is easy to understand. It's probably on the same difficulty as being able to notice that if you increase the number of sides of a polygon to infinity, your shape starts to look like a circle.

The reason we teach concepts of calculus at a higher level than basic arithmetic is because of the computation.


I guess, but what about these "Further maths iGCSEs", why instead of limiting teaching slightly higher math to a select few, we can just make the basic concept part of the normal GCSE course,
Reply 17
Original post by tohaaaa
I guess, but what about these "Further maths iGCSEs", why instead of limiting teaching slightly higher math to a select few, we can just make the basic concept part of the normal GCSE course,

TBH you can start teaching concepts of calculus at primary school, at least without a lot of computation.

Although I'd love to see calculus taught to all GCSE students, I doubt that is going to happen any time soon.
Reply 18
Original post by 0x2a
TBH you can start teaching concepts of calculus at primary school, at least without a lot of computation.

Although I'd love to see calculus taught to all GCSE students, I doubt that is going to happen any time soon.


Maybe not primary school.

But why do you doubt it would happen?
Reply 19
Original post by tohaaaa
Maybe not primary school.

But why do you doubt it would happen?

We probably won't start teaching calculus at primary schools before 3000... but basic concepts of calculus can easily be understood at this age. Young child already know that if you zoom really close into a curved line it starts to look straight (Differentiation).

I mean there is basic differentiation and integration in additional maths (FSMQ), that won't be too hard for higher-tier GCSE students at least. But if you look at the standards of GCSE maths papers these days, it's pretty shocking. If we're going to teach any serious calculus at GCSE, then we're going to have to reform primary school education too.

Also as time passes other topics might become more significant. For example a maths curriculum based on problem solving, and not doing tons of arithmetic, might become popular.

I think we've gone off tangent a bit too much...
(edited 10 years ago)

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