Limits of a function

straight_As

Hey there

I have my LC next week and was just looking over some questions on arithmetico geometric series when I came across a limit that I couldn't quite get my head around.

Basically

lim(n -> infinity) of n(x^n)/(1-x)... |x| < 1

According to the answer at the back of the book, it appears to tend to zero, but I'm not quite sure why.

x^n will tend towards zero, I know, but what about n/(1-x)? Does that not go to an infinitely big number.

So here's the crux of my problem; (infinitely big) x (infinitely small) = ? (and why?)

For LC level do I just have to accept that once one part of the function tends towards zero, the entire function must?

Cheers in advance

Michael Collins

A bit of a coincidence that I just posted a question on that very topic a few weeks ago!

http://www.boards.ie/vbulletin/showthread.php?t=2055912965

straight_As wrote: »

...For LC level do I just have to accept that once one part of the function tends towards zero, the entire function must?

Cheers in advance

This is not true in general. If the rest of the function goes to some finite limit and some factor goes to 0, then yes it will be true. But, as you have pointed out, while one factor of the term in question goes to 0, the other gets infinitely big - this just isn't doable to any real extent on the LC, so yes you do just have to accept it I'm afraid. But check out the thread I've linked to above, there's some good motivations in that which should help convince you.

If you are interested, it doesn't take much extra knowledge to calculate limits like this one, they are called "indeterminate forms" i.e. limits where you seem to get terms like:

[latex] \displaystyle 0 \times \infty [/latex] or [latex] \displaystyle \frac{0}{0} [/latex] or [latex] \displaystyle \frac{\infty}{\infty} [/latex] (and others).

The method for evaluating these is called L'Hôpital's rule.

In this example we have

[latex] \displaystyle \lim_{n \to \infty} \frac{nx^{n+1}}{1-x} = \frac{\infty. 0}{1-x} [/latex]

we rewrite this as:

[latex] \displaystyle \lim_{n \to \infty} \frac{1}{1-x} \frac{n}{x^{-(n+1)}} = \frac{1}{1-x}\lim_{n \to \infty} \frac{n}{x^{-(n+1)}} = \frac{1}{1-x}\frac{\infty}{\infty} [/latex]

so apply L'Hôpital's rule to the [latex] \frac{\infty}{\infty} [/latex] term by differentiating the numerator and denominator with respect to n:

[latex] \displaystyle \lim_{n \to \infty} \frac{n}{x^{-(n+1)}} = \lim_{n \to \infty} \frac{1}{-(n+1)x^{-(n+2)}} = 0 [/latex]

with the last equality due to the fact that the denominater gets infinitely big while the numerator is now a constant (=1).

Try this yourself with

[latex] \displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} [/latex]

(people may complain here because L'Hôpital's rule is often used to prove the derivative of sin and so you have circular logic, but don't worry about that just yet!)

straight_As

Cheers, that helps a lot

Re: That sinx question, is this ok?

lim (x -> 0) [sin(x)/x]

Apply l'Hopital's Rule, therefore

lim (x -> 0) [sin(x)/x] = lim (x -> 0) [(d(sin(x))/dx)/(d(x)/dx)]

= lim (x -> 0) [cos(x)/1]

= 1

That's pretty neat! Would that be an acceptable way to answer those limit questions in the LC?

equivariant

straight_As wrote: »

Cheers, that helps a lot

Re: That sinx question, is this ok?

lim (x -> 0) [sin(x)/x]

Apply l'Hopital's Rule, therefore

lim (x -> 0) [sin(x)/x] = lim (x -> 0) [(d(sin(x))/dx)/(d(x)/dx)]

= lim (x -> 0) [cos(x)/1]

= 1

That's pretty neat! Would that be an acceptable way to answer those limit questions in the LC?

I hope not. As Michael Collins pointed out earlier, this argument employs circular reasoning (i.e. it assumes the very thing that it is trying to prove). There is a discussion of this limit in a previous thread
http://boards.ie/vbulletin/showthread.php?t=2055891303

Michael Collins

straight_As wrote: »

Cheers, that helps a lot

Re: That sinx question, is this ok?

lim (x -> 0) [sin(x)/x]

Apply l'Hopital's Rule, therefore

lim (x -> 0) [sin(x)/x] = lim (x -> 0) [(d(sin(x))/dx)/(d(x)/dx)]

= lim (x -> 0) [cos(x)/1]

= 1

Yep, perfect!

That's pretty neat! Would that be an acceptable way to answer those limit questions in the LC?

equivariant wrote: »

I hope not. As Michael Collins pointed out earlier, this argument employs circular reasoning (i.e. it assumes the very thing that it is trying to prove). There is a discussion of this limit in a previous thread
http://boards.ie/vbulletin/showthread.php?t=2055891303

Actually, it appears to be!

Believe it or not L'Hôpital's rule has appeared as part of the marking scheme, take LC Higher Level Maths 2005 Paper 2, Q4(a), the marking scheme shows a calculation using L'Hôpital's rule:

http://www.examinations.ie/archive/markingschemes/2005/LC003ALP2EV.pdf

I don't know why, I suppose because L'Hôpital's rule is quite straightforward to use the State Examinations Commission considerd that some teachers may have taught their students this method, or else it was actually on a previous version of the syllabus - any stalwarts here who can remember the syllabus before the last change (in 1994 I think?).

Fremen

equivariant wrote: »

I hope not. As Michael Collins pointed out earlier, this argument employs circular reasoning (i.e. it assumes the very thing that it is trying to prove). There is a discussion of this limit in a previous thread
http://boards.ie/vbulletin/showthread.php?t=2055891303

That's really not the case anymore in modern mathematics. For example, the complex analytic definitions of sin and cos are

[latex] \displaystyle \sin(z) = \frac{e^{iz} - e^{-iz}}{2i}[/latex]


[latex] \displaystyle \cos(z) = \frac{e^{iz} + e^{-iz}}{2}[/latex]

Differentiate them and the usual relations pop right out.

I agree it's circular when you restrict yourself to leaving cert methods, but then again if you did that, you wouldn't be using l'hopital to begin with.

CJC86

Michael Collins wrote: »

so apply L'Hôpital's rule to the [latex] \frac{\infty}{\infty} [/latex] term by differentiating the numerator and denominator with respect to n:

[latex] \displaystyle \lim_{n \to \infty} \frac{n}{x^{-(n+1)}} = \lim_{n \to \infty} \frac{1}{-(n+1)x^{-(n+2)}} = 0 [/latex]

with the last equality due to the fact that the denominater gets infinitely big while the numerator is now a constant (=1).

Definitely something up with this here. You can't differentiate with respect to n (well, you could take a formal derivative, but it's pretty meaningless).

Probably best to leave an x in the numerator, while taking x^n down. Then taking both derivatives with respect to x, we get:

[latex] \displaystyle \lim_{n \to \infty} \frac{nx}{x^{-n}} = \lim_{n \to \infty} \frac{n}{-nx^{-n-1}} = \lim_{n \to \infty} \frac{1}{x^{-n-1}} = \lim_{n \to \infty}x^{n+1} = 0 [/latex]

I can't think of a way to do the problem without using l'Hopital at the moment. Actually, could probably do it easily with the epsilon definition of limits, but that's not on the leaving cert syllabus either.

Michael Collins

CJC86 wrote: »

Definitely something up with this here. You can't differentiate with respect to n (well, you could take a formal derivative, but it's pretty meaningless).

Probably best to leave an x in the numerator, while taking x^n down. Then taking both derivatives with respect to x, we get:

[latex] \displaystyle \lim_{n \to \infty} \frac{nx}{x^{-n}} = \lim_{n \to \infty} \frac{n}{-nx^{-n-1}} = \lim_{n \to \infty} \frac{1}{x^{-n-1}} = \lim_{n \to \infty}x^{n+1} = 0 [/latex]

I can't think of a way to do the problem without using l'Hopital at the moment. Actually, could probably do it easily with the epsilon definition of limits, but that's not on the leaving cert syllabus either.

:eek: Opps. I ignored the fact that n is an integer! Is it legitimate to differentiate with any old variable in applying l'Hopital's Rule, I always thought it had to be the variable you're taking the limit with respect to?

equivariant

Fremen wrote: »

That's really not the case anymore in modern mathematics. For example, the complex analytic definitions of sin and cos are

[latex] \displaystyle \sin(z) = \frac{e^{iz} - e^{-iz}}{2i}[/latex]


[latex] \displaystyle \cos(z) = \frac{e^{iz} + e^{-iz}}{2}[/latex]

Differentiate them and the usual relations pop right out.

I agree it's circular when you restrict yourself to leaving cert methods, but then again if you did that, you wouldn't be using l'hopital to begin with.

To rigourously define and then calculate the derivative of the complex exponential function uses mathematics far more sophisticated than l'Hopital and/or basic trig. So to appeal to this argument as a rigourous derivation of the limit is not really acceptable.

In the thread that I linked earlier, MathsManiac outlined a rigourous derivation of that limit, but it is fairly involved and is beyond the scope of leaving cert mathematics I think (at least in my opinion).

equivariant

Michael Collins wrote: »

A bit of a coincidence that I just posted a question on that very topic a few weeks ago!

http://www.boards.ie/vbulletin/showthread.php?t=2055912965



This is not true in general. If the rest of the function goes to some finite limit and some factor goes to 0, then yes it will be true. But, as you have pointed out, while one factor of the term in question goes to 0, the other gets infinitely big - this just isn't doable to any real extent on the LC, so yes you do just have to accept it I'm afraid. But check out the thread I've linked to above, there's some good motivations in that which should help convince you.

If you are interested, it doesn't take much extra knowledge to calculate limits like this one, they are called "indeterminate forms" i.e. limits where you seem to get terms like:

[latex] \displaystyle 0 \times \infty [/latex] or [latex] \displaystyle \frac{0}{0} [/latex] or [latex] \displaystyle \frac{\infty}{\infty} [/latex] (and others).

The method for evaluating these is called L'Hôpital's rule.

In this example we have

[latex] \displaystyle \lim_{n \to \infty} \frac{nx^{n+1}}{1-x} = \frac{\infty. 0}{1-x} [/latex]

we rewrite this as:

[latex] \displaystyle \lim_{n \to \infty} \frac{1}{1-x} \frac{n}{x^{-(n+1)}} = \frac{1}{1-x}\lim_{n \to \infty} \frac{n}{x^{-(n+1)}} = \frac{1}{1-x}\frac{\infty}{\infty} [/latex]

so apply L'Hôpital's rule to the [latex] \frac{\infty}{\infty} [/latex] term by differentiating the numerator and denominator with respect to n:

[latex] \displaystyle \lim_{n \to \infty} \frac{n}{x^{-(n+1)}} = \lim_{n \to \infty} \frac{1}{-(n+1)x^{-(n+2)}} = 0 [/latex]

with the last equality due to the fact that the denominater gets infinitely big while the numerator is now a constant (=1).

Try this yourself with

[latex] \displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} [/latex]

(people may complain here because L'Hôpital's rule is often used to prove the derivative of sin and so you have circular logic, but don't worry about that just yet!)

You have made an error with the differentiation here, as you differentiated with respect to n in the numerator and with respect to x in the denominator.

The issue of whether or not n is an integer is not really an issue here as, for positive values of x we can calculate as if n is a real number and then specialise to the case of n being an integer if necessary.

Anyway, here is a proof that uses only the squeeze theorem (or sandwich theorem if you prefer) - no need for l'Hopital.

Choose N so that [latex] |x| < \frac{N}{N + 1} [/latex]

We can do this since [latex] \lim_{m\rightarrow \infty \frac{m}{m + 1} = 1[/latex] and by assumption [latex] |x| <1 [/latex]

Now let [latex] y = \frac{N + 1}{N}x [/latex]. Clearly [latex]|y| <1[/latex]

Now, if [latex]n>N[/latex] we have [latex]
|nx^n| = |x^N||\frac{n}{n - 1}x||\frac{n - 1}{n - 2}x| \dots |\frac{N + 1}{N} x|N \leq |x^N| |y^{n-N}| N[/latex] since [latex] |\frac{k + 1}{k} x| \leq |y| [/latex] for all [latex] k \geq N[/latex]

So, for all [latex] n>N [/latex], we have [latex] 0 \leq |nx^n| \leq N|\frac{x^N}{y^N}| |y^n|[/latex].

Now [latex] \lim_{n\rightarrow \infty} |y^n| = 0 [/latex] since [latex]|y| <1 [/latex]. So by the squeeze theorem, [latex] \lim_{n\rightarrow \infty} nx^n = 0 [/latex]

RoundTower

I sat my LC in 2002 and we were taught l'Hopital's rule. Given the teacher it's entirely possible it was no longer on the syllabus at that point though!

Michael Collins

equivariant wrote: »

You have made an error with the differentiation here, as you differentiated with respect to n in the numerator and with respect to x in the denominator.

The issue of whether or not n is an integer is not really an issue here as, for positive values of x we can calculate as if n is a real number and then specialise to the case of n being an integer if necessary.

Anyway, here is a proof that uses only the squeeze theorem (or sandwich theorem if you prefer) - no need for l'Hopital.

Choose N so that [latex] |x| < \frac{N}{N + 1} [/latex]

We can do this since [latex] \lim_{m\rightarrow \infty \frac{m}{m + 1} = 1[/latex] and by assumption [latex] |x| <1 [/latex]

Now let [latex] y = \frac{N + 1}{N}x [/latex]. Clearly [latex]|y| <1[/latex]

Now, if [latex]n>N[/latex] we have [latex]
|nx^n| = |x^N||\frac{n}{n - 1}x||\frac{n - 1}{n - 2}x| \dots |\frac{N + 1}{N} x|N \leq |x^N| |y^{n-N}| N[/latex] since [latex] |\frac{k + 1}{k} x| \leq |y| [/latex] for all [latex] k \geq N[/latex]

So, for all [latex] n>N [/latex], we have [latex] 0 \leq |nx^n| \leq N|\frac{x^N}{y^N}| |y^n|[/latex].

Now [latex] \lim_{n\rightarrow \infty} |y^n| = 0 [/latex] since [latex]|y| <1 [/latex]. So by the squeeze theorem, [latex] \lim_{n\rightarrow \infty} nx^n = 0 [/latex]

Right. So you'd say that there's nothing wrong with

[latex] \displaystyle \lim_{n \to \infty} \frac{n}{x^{-(n+1)}} = \lim_{n \to \infty} \frac{\frac{d}{dn}\; n}{\frac{d}{dn}\; x^{-(n+1)}} = \lim_{n \to \infty} \frac{1}{-\ln(x)\;x^{-(n+1)}} = 0 [/latex]

equivariant

Michael Collins wrote: »

Right. So you'd say that there's nothing wrong with

[latex] \displaystyle \lim_{n \to \infty} \frac{n}{x^{-(n+1)}} = \lim_{n \to \infty} \frac{\frac{d}{dn}\; n}{\frac{d}{dn}\; x^{-(n+1)}} = \lim_{n \to \infty} \frac{1}{-\ln(x)\;x^{-(n+1)}} = 0 [/latex]

That works I think.

MathsManiac

equivariant wrote: »

...

Choose N so that [latex] |x| < \frac{N}{N + 1} [/latex]

We can do this since [latex] \lim_{m\rightarrow \infty \frac{m}{m + 1} = 1[/latex] and by assumption [latex] |x| <1 [/latex]

Now let [latex] y = \frac{N + 1}{N}x [/latex]. Clearly [latex]|y| <1[/latex]

Now, if [latex]n>N[/latex] we have [latex]
|nx^n| = |x^N||\frac{n}{n - 1}x||\frac{n - 1}{n - 2}x| \dots |\frac{N + 1}{N} x|N \leq |x^N| |y^{n-N}| N[/latex] since [latex] |\frac{k + 1}{k} x| \leq |y| [/latex] for all [latex] k \geq N[/latex]

So, for all [latex] n>N [/latex], we have [latex] 0 \leq |nx^n| \leq N|\frac{x^N}{y^N}| |y^n|[/latex].

Now [latex] \lim_{n\rightarrow \infty} |y^n| = 0 [/latex] since [latex]|y| <1 [/latex]. So by the squeeze theorem, [latex] \lim_{n\rightarrow \infty} nx^n = 0 [/latex]

Excellent job!

(By the way, I've also heard that theorem referred to as the fly-swatter theorem, which is a nice image. No matter how the fly behaves, if he's always confined to the region between the swatter and the wall, he's doomed!)

ray giraffe

straight_As wrote: »

Hey there

I have my LC next week and was just looking over some questions on arithmetico geometric series when I came across a limit that I couldn't quite get my head around.

Basically

lim(n -> infinity) of n(x^n)/(1-x)... |x| < 1

According to the answer at the back of the book, it appears to tend to zero, but I'm not quite sure why.

x^n will tend towards zero, I know, but what about n/(1-x)? Does that not go to an infinitely big number.

So here's the crux of my problem; (infinitely big) x (infinitely small) = ? (and why?)

For LC level do I just have to accept that once one part of the function tends towards zero, the entire function must?

Cheers in advance

The following method has the advantage that the Ratio Test is on the Leaving Cert course.

Let the expression be [latex] \displaystyle \lim_{n \to \infty} a_n[/latex].

Note that by the Ratio Test [latex] \displaystyle \sum_{n=1}^{\infty} a_n[/latex] converges.

Then recall that if [latex] \displaystyle \sum_{n=1}^\infty a_n[/latex] converges, then [latex]\lim_{n \to \infty} a_n = 0 [/latex]. We are finished!

Note: A 2-line proof of the last fact is under "Limit Manipulation" in http://en.wikipedia.org/wiki/N-th_term_test#Proofs