## How is l'hopital useful here?

**Moderators:** gmalivuk, Moderators General, Prelates

### How is l'hopital useful here?

Alright, so this question is 1 part me figuring out how to tex-ify this, 1 part checking my reasoning on something is correct, and one part potentially help on how to attack part of a homework problem, although I haven't spent long enough on it to consider myself 'stuck' yet.

So, I'm given the following function

[math]f(x) = \left \{ \begin{array}{lrr} e^{\frac{-1}{x^2}} & for & x \ne 0\\ 0 & for & x=0 \end{array} \right.[/math]

(Latex questions: Is there a better way to avoid having "for x" run together other than giving "for" it's own column? Can I make the exponent of e look nicer in it's present form?)

And I need to show that [imath]f'(0)=0[/imath] and [imath]f^{(n)}(0)=0[/imath]. ([imath]f^{(n)}[/imath]means the n-th derivative, not some sort of composition of the function.) It was mentioned by the professor that we were welcome to use l'hopitals rule to make things easier if we wished.

By my reckoning, the derivative is [math]f'(x)= \frac{2e^{\frac{-1}{x^2}}}{x^3} =\frac{2}{x^3e^{x^{-2}}} =\frac{2}{x^3[1+x^{-2}+\frac{(x^{-2})^2}{2!}+\ldots]}=\frac{2}{[x^3+x^{1}+\frac{x^{-1}}{2!}+\ldots]}[/math]

And thus

[math]f'(0)=\frac{2}{[0+0+\infty+\infty+\ldots]} = 0[/math]

Is the above conlusion valid, or is there trouble lurking in that infinite sum of infinities that might somehow cause that to not necessarily be 0? (I appreciate I should probably have a lim x->0 step but that's probably besides the point.)

I see that while I could have tried to use l'hopitals rule on that first form of f'(x) to find f'(0) since it'd be of the form 0/0, the derivative of the numerator would be worse, not better, unless I'm making a stupid mistake there.

From here, what would be the best way to show [imath]f^{(n)}(0)=0[/imath]? I imagine that it'll involve my having shown that the [imath]f^{(n-1)}(0)[/imath] term is 0 and as such theres probably some induction to be done, but I'm not sure if I can directly work with [imath]f^{(n)}(0)[/imath] somehow, or if I'll need to figure out what [imath]f^{(n)}(x)[/imath] looks like. Also, I'd like to think to think l'hopitals rule was mentioned for a good reason and that theres a way to put it to work without making the situation seem worse, but I don't see where...

So, I'm given the following function

[math]f(x) = \left \{ \begin{array}{lrr} e^{\frac{-1}{x^2}} & for & x \ne 0\\ 0 & for & x=0 \end{array} \right.[/math]

(Latex questions: Is there a better way to avoid having "for x" run together other than giving "for" it's own column? Can I make the exponent of e look nicer in it's present form?)

And I need to show that [imath]f'(0)=0[/imath] and [imath]f^{(n)}(0)=0[/imath]. ([imath]f^{(n)}[/imath]means the n-th derivative, not some sort of composition of the function.) It was mentioned by the professor that we were welcome to use l'hopitals rule to make things easier if we wished.

By my reckoning, the derivative is [math]f'(x)= \frac{2e^{\frac{-1}{x^2}}}{x^3} =\frac{2}{x^3e^{x^{-2}}} =\frac{2}{x^3[1+x^{-2}+\frac{(x^{-2})^2}{2!}+\ldots]}=\frac{2}{[x^3+x^{1}+\frac{x^{-1}}{2!}+\ldots]}[/math]

And thus

[math]f'(0)=\frac{2}{[0+0+\infty+\infty+\ldots]} = 0[/math]

Is the above conlusion valid, or is there trouble lurking in that infinite sum of infinities that might somehow cause that to not necessarily be 0? (I appreciate I should probably have a lim x->0 step but that's probably besides the point.)

I see that while I could have tried to use l'hopitals rule on that first form of f'(x) to find f'(0) since it'd be of the form 0/0, the derivative of the numerator would be worse, not better, unless I'm making a stupid mistake there.

From here, what would be the best way to show [imath]f^{(n)}(0)=0[/imath]? I imagine that it'll involve my having shown that the [imath]f^{(n-1)}(0)[/imath] term is 0 and as such theres probably some induction to be done, but I'm not sure if I can directly work with [imath]f^{(n)}(0)[/imath] somehow, or if I'll need to figure out what [imath]f^{(n)}(x)[/imath] looks like. Also, I'd like to think to think l'hopitals rule was mentioned for a good reason and that theres a way to put it to work without making the situation seem worse, but I don't see where...

### Re: How is l'hopital useful here?

let g(x) = exp(-1/[x^2])=1/(exp(1/x^2)

lim+ (g) x -> 0 = 1/ lim+(exp(1/x^2)) x -> 0 = 0

lim- (g) x -> 0 = 1/lim-(exp(1/x^2)) x -> 0 = 0

so f is continuous at 0 (and along all of R), and the function is also sufficiently smooth.

An important thing to think about here is that g'(x), as you noticed, is undefined at 0, remember that

lim(f(x))= infinity is actually just a fancy notation to describe the behaviour of a function who's limit doesn't actually exist. As a result, if a function's limit goes to infinity at a point, the function is discontinuous at that point and the derivative doesn't exist. You can't divide by infinity.

So if that's the case, then how can the nth derivative of f(x) be 0 for all integers n? (pro-tip, the integrals are also zero at f(x)=0) what simple rule can you use to show this? You are right that the derivative of g(x) is 2/((x^3)*(e^-1/x^2)); however, the discontinuity at x=0 is from the 1/(x^3) part, not the exponential.

lim+ (g) x -> 0 = 1/ lim+(exp(1/x^2)) x -> 0 = 0

lim- (g) x -> 0 = 1/lim-(exp(1/x^2)) x -> 0 = 0

so f is continuous at 0 (and along all of R), and the function is also sufficiently smooth.

An important thing to think about here is that g'(x), as you noticed, is undefined at 0, remember that

lim(f(x))= infinity is actually just a fancy notation to describe the behaviour of a function who's limit doesn't actually exist. As a result, if a function's limit goes to infinity at a point, the function is discontinuous at that point and the derivative doesn't exist. You can't divide by infinity.

So if that's the case, then how can the nth derivative of f(x) be 0 for all integers n? (pro-tip, the integrals are also zero at f(x)=0) what simple rule can you use to show this? You are right that the derivative of g(x) is 2/((x^3)*(e^-1/x^2)); however, the discontinuity at x=0 is from the 1/(x^3) part, not the exponential.

- Talith
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### Re: How is l'hopital useful here?

I believe that for this question you were probably meant to use the definition of derivative at a point. That is, f'(x) at x=a is definied to be (whenever this limit is well defined) [math]f'(x)|_{x=a}=\lim_{h\rightarrow 0} \frac{f(a+h)-f(a)}{h}[/math]The way you've gone about answering the question doesn't really work because how do you know what the derivative of the function is? It's a piece-wise function which I'm pretty sure you've never formally differentiated before. When evaluating the limit, try to remember that you can't just do normal arithmetic on infinities (explicit the way you have, or implicit in the sense of 1/0) so use l'hopitals rule (or use the definition of a limit if you have some time on your hands).

### Re: How is l'hopital useful here?

Talith wrote:I believe that for this question you were probably meant to use the definition of derivative at a point. That is, f'(x) at x=a is definied to be (whenever this limit is well defined) [math]f'(x)|_{x=a}=\lim_{h\rightarrow 0} \frac{f(a+h)-f(a)}{h}[/math]

I actually did that initially, but since the a I'm interested in is a=0, the limit becomes [math]f'(x)|_{x=0}=\lim_{h\rightarrow 0} \frac{f(0+h)-f(0)}{h} = \lim_{h\rightarrow 0} \frac{f(h)-0}{h}[/math]

And that looks to me like I'd have exactly the same issue when attempting to apply l'hopitals to it, as I'd have to differentiate the numerator, which would potentially be defeating the purpose as that's the derivative I'm trying to find in the first place, except with an h in place of the x.

(The surrounding course context is on DE's and fourier series, and the problem I believe is meant to demonstrate that just because you can compute a series doesn't mean it will converge to the original function [since the result being proven would in turn imply that the taylor series about x=0 would be 0 everywhere, not just at x=0])

- Yakk
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### Re: How is l'hopital useful here?

For l'Hopital to hold, your function must be differentiable.

Do you know anything about "stitching" functions together? Are you allowed to hand-wave it? What is the level of rigor expected in your course?

You need to prove the result for all n. With the benefit of hindsight, I might ask you what lim(x->0) of t_n(x) := e^(-1/x) / x^n is?

Do you know anything about "stitching" functions together? Are you allowed to hand-wave it? What is the level of rigor expected in your course?

You need to prove the result for all n. With the benefit of hindsight, I might ask you what lim(x->0) of t_n(x) := e^(-1/x) / x^n is?

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

- phlip
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### Re: How is l'hopital useful here?

But since you know that h is nonzero, you can differentiate to find your f'(h) in the normal way, with the chain rule and friends... since you know that a neighbourhood of h is entirely within one piece of your piecewise function, you only have to differentiate that piece. As opposed to finding f'(0), for which you have to examine multiple pieces, as it's a boundary point.

That is, you have g(x) = e^(-x^-2), and f(x) = g(x) for x != 0, 0 for x = 0. Then you can find g'(x) easily enough, and then f'(x) = g'(x) for x != 0, <complicated thing you're trying to figure out> for x = 0. As long as, in the process of calculating f'(0), you only have to evaluate f'(h) for nonzero values of h, you're golden.

That is, you have g(x) = e^(-x^-2), and f(x) = g(x) for x != 0, 0 for x = 0. Then you can find g'(x) easily enough, and then f'(x) = g'(x) for x != 0, <complicated thing you're trying to figure out> for x = 0. As long as, in the process of calculating f'(0), you only have to evaluate f'(h) for nonzero values of h, you're golden.

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### Re: How is l'hopital useful here?

You're still missing a pretty obvious trick, the limits need only be used to prove continuity, because of how the function is defined, I'm trying to not just give away the trick here, but keep in mind the function g(x) = exp(-1/(x^2)) is only once-differentiable, and the question specifically asks why the function f(x) = {g(x) for x != 0; 0 for x = 0} has a derivative equal to 0; going farther than than that, it asks why the function's nth derivative, where n can be any number1, 2, 3, ..., is equal to zero. Since g'(x) is discontinuous at 0, can g''(x) exist at the point x=0?

It might also be helpful to think of it this way, if f'(x)=g'(x), then can f''(0) exist, or can f'(0) exist; you've already seen that g'(x) is discontinuous at 0, and you know something about f(x) makes it continuous at 0, and infinitely differentiable; what kinds of things do you know have this property?

It might also be helpful to think of it this way, if f'(x)=g'(x), then can f''(0) exist, or can f'(0) exist; you've already seen that g'(x) is discontinuous at 0, and you know something about f(x) makes it continuous at 0, and infinitely differentiable; what kinds of things do you know have this property?

### Re: How is l'hopital useful here?

Yakk wrote:For l'Hopital to hold, your function must be differentiable.

Do you know anything about "stitching" functions together? Are you allowed to hand-wave it? What is the level of rigor expected in your course?

You need to prove the result for all n. With the benefit of hindsight, I might ask you what lim(x->0) of t_n(x) := e^(-1/x) / x^n is?

My interpretation of stiching functions together is that it's basicly just when you've got a piecewise function, but they're equal at the boundaries (e.g. |x| is two lines 'stiched' together at 0. If theres a more rigourous definition, I don't know it.) Theres lots and lots of handwaving in the course, since apparently one needs a background in analysis for most of the proofs, which the majority of the class lacks (including myself). I wouldn't consider it a particularly rigour demanding course. We've needed to use induction to prove the appropriate form of c

_{n}when dealing with infinite series instead of just sortof looking at the first few results and going 'yup, it's this.', but not much more.

With regards to the limit you mention, just considering the n=0 case, the limit must be 0, but it's not obvious to me how to prove that, since that is clearly not differentiable at x=0 which breaks my main method of limit evaluating. edit2: Actually, that one wouldn't even necessarily be 0 in the n=0 case, since coming from the other side you'd have infinity, that is, that limit doesn't exist.

edit:

vilidice wrote: Since g'(x) is discontinuous at 0, can g''(x) exist at the point x=0?

I'm inclined to say no.

It might also be helpful to think of it this way, if f'(x)=g'(x), then can f''(0) exist, or can f'(0) exist; you've already seen that g'(x) is discontinuous at 0, and you know something about f(x) makes it continuous at 0, and infinitely differentiable; what kinds of things do you know have this property?

Constants, and for that matter polynomials are all continuous at 0 and infinitely differentiable. I would imagine theres other stuff too that I'm not thinking of.

The derivative of a piecewise function strikes me as something that itself wants to be piecewise as well. Is that a line of thinking I should be pursuing, or can I actually do anything with it?

Last edited by Dopefish on Thu Feb 17, 2011 10:56 pm UTC, edited 1 time in total.

### Re: How is l'hopital useful here?

Dopefish wrote:(Latex questions: Is there a better way to avoid having "for x" run together other than giving "for" its own column? Can I make the exponent of e look nicer in its present form?)

My answers to those questions (others may have different tastes):

**Spoiler:**

### Re: How is l'hopital useful here?

If you put the two ideas from your last post together, and do a little bit of thinking in through the boundaries you should see what's happening there. Keep in mind the type of discontinuity the derivative of g(x) has at x=0, and how changing the function at that one point makes the second derivative exist.

### Re: How is l'hopital useful here?

I observe that the discontinuity in the original function is just a hole, but the limit as the surrounding function approaches 0 from either side exists and is 0, hence defining the function to be 0 at 0 makes it continuous and nice. Also, g(x) is even, while g'(x) isn't, which causes the limit at 0 to not exist (of course, the limit could be made to exist without the function being even, but evenness looks like it'd be a sufficient condition since the function is defined on either side of 0). Possibly useful, possibly not directly relevent though.

I'm not entirely sure what I'm allowed to do that would help. I don't think I can define anything new to make things behave better, since that would basicly be changing the problem. (Otherwise I could just define f'(x) and f^n(x) to be 0 and say they're 0 there because I defined them as such. ) From the sounds of things, there is clearly some trick that makes it all work out that I'll promptly consider myself an idiot for not immediately recognising, but for now I'm not seeing what I could do...

I also still haven't seen anywhere where l'hopital is useful, since if I use it on something involving that exponential, it spawns an x^-3 which worsens things rather than makes them better.

I'm not entirely sure what I'm allowed to do that would help. I don't think I can define anything new to make things behave better, since that would basicly be changing the problem. (Otherwise I could just define f'(x) and f^n(x) to be 0 and say they're 0 there because I defined them as such. ) From the sounds of things, there is clearly some trick that makes it all work out that I'll promptly consider myself an idiot for not immediately recognising, but for now I'm not seeing what I could do...

I also still haven't seen anywhere where l'hopital is useful, since if I use it on something involving that exponential, it spawns an x^-3 which worsens things rather than makes them better.

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### Re: How is l'hopital useful here?

Dopefish wrote:Also, g(x) is even, while g'(x) isn't, which causes the limit at 0 to not exist

No, the limit at 0 of g'(x) does exist. Note that the derivative of an even function is an odd function... and the limit at 0 of an odd function can exist if it's 0. The challenge is to prove that the limit of g'(x) at 0 is 0 (and thus that f'(0) = 0).

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### Re: How is l'hopital useful here?

phlip wrote:Dopefish wrote:Also, g(x) is even, while g'(x) isn't, which causes the limit at 0 to not exist

No, the limit at 0 of g'(x) does exist. Note that the derivative of an even function is an odd function... and the limit at 0 of an odd function can exist if it's 0. The challenge is to prove that the limit of g'(x) at 0 is 0 (and thus that f'(0) = 0).

I actually don't think I knew that the derivative of an even was necessarily odd. Yay for learning.

Hmm...I guess it should have been obvious that the limit must've existed given the quesiton is basicly saying I need to show it does, although graphing what I believe to be g'(x) show it going to positive infinity on the negative side of 0, and going to 0 from the positive side. Probably a sign that I haven't entered things properly...and indeed I was missing a parenthesis, now properly creating an odd function that wants to be 0 at 0. Proving it to be as such remains a challenge, but at least now my clalculator isn't suggesting that it's an impossible challenge.

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### Re: How is l'hopital useful here?

Dopefish wrote:I actually don't think I knew that the derivative of an even was necessarily odd. Yay for learning.

Yep, and vice-versa. Flows straight from the chain rule: d/dx f(-x) = -f'(-x), so if f(x) = f(-x) then f'(x) = -f'(-x).

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### Re: How is l'hopital useful here?

Dopefish wrote:

[math]f(x) = \left \{ \begin{array}{lrr} e^{\frac{-1}{x^2}} & for & x \ne 0\\ 0 & for & x=0 \end{array} \right.[/math]

(Latex questions: Is there a better way to avoid having "for x" run together other than giving "for" it's own column? Can I make the exponent of e look nicer in it's present form?)

\mbox was mentioned earlier. I also like using \text{ } since it's easier for me to remember. On another note instead of using an array there is the \align or \align* environments which work much the same but are "smart" and you don't have to specify the number of columns or the alignment type in each column. More suited to this is the \cases environment. Which works much the same as \align but adds in the \left brace for you. So not a big difference just saves a few keystrokes.

I played with the exponent a little bit.... You can of course use \displaystyle to make it bigger. which looks like [imath]e^{\displaystyle\frac{-1}{x^2}}[/imath] so the "exponent part" doesn't look too superscripty anymore. In the version below I put made the exponent on "e" be a \phantom{.} with the actual exponent a superscript of that. It might be what you're looking for ... although using the / instead of \frac{}{} is probably the way most people would reformat that... or do a [imath]e^{-\left(\frac{1}{x}\right)^2}[/imath] instead to get that extra power out of the denominator, which isn't perfect but might help a little. Anyway here is \cases in action.

[math]f(x) = \begin{cases} e^{\phantom{.}^{\displaystyle\frac{-1}{x^2}}} &\text{ for } x\neq 0 \\ 0 &\text{ for } x=0 \end{cases}[/math]

### Re: How is l'hopital useful here?

Thanks for the latex help, and guidence on the question. Still learning the various tex tricks around and often needing to look up appropriate commands whenever I need something, but it's coming along, and it's ever so pretty.

The professors solution has since been posted, and basicly involves making a substition l=1/x (which I could've sworn I'd tried at some stage without it working out, but ah well), then showing that of g'(x) goes to 0 in the limit of x->0 (i.e. l->infinity), and since it's continous thus the derivative is also 0. This was done for the first, second, 3rd, and 4th derivatives using basicly the same procedure, and then it was stated that 'clearly' all higher derivatives sums of a*e^(1/x^2)/x^k for some constants a and k, each of which were obviously 0 in the limit, and thus f

In a distantly related question (in that it involves me not necessarily approving of my professors answer, but I don't think it warrents it's own thread), on a previous assignment, we needed to differentiate [math]c_{2n}(r)=\frac{(-1)^n}{(r+2)^2(r+4)^2\cdots(r+2n)^2}[/math] with respect to r, in order to show that [math]c'_{2n}(0)=\frac{(-1)^{n+1}H_n}{2^{2n}(n!)^2}[/math]

My solution involved expanding the denominator of [imath]c_{2n}(r)[/imath] into a polynomial of degree n squared with undetermined coefficients b

While that gives the right result for r=0, it strikes me as if that differentiation is neglecting the product rule, and thus shouldn't work. Am I missing something that does make it work out as he did it, or is his solution flawed?

The professors solution has since been posted, and basicly involves making a substition l=1/x (which I could've sworn I'd tried at some stage without it working out, but ah well), then showing that of g'(x) goes to 0 in the limit of x->0 (i.e. l->infinity), and since it's continous thus the derivative is also 0. This was done for the first, second, 3rd, and 4th derivatives using basicly the same procedure, and then it was stated that 'clearly' all higher derivatives sums of a*e^(1/x^2)/x^k for some constants a and k, each of which were obviously 0 in the limit, and thus f

^{n}(0)=0 for all n. Although it's actually basicly what I did in showing it, it still feels terribly not rigourous since it was only shown for n=1,2,3,4. Of course, it's obvious from chain rule/product rule that the higer derivatives will be of that form, and that they'll be 0, but obvious != proof to me.In a distantly related question (in that it involves me not necessarily approving of my professors answer, but I don't think it warrents it's own thread), on a previous assignment, we needed to differentiate [math]c_{2n}(r)=\frac{(-1)^n}{(r+2)^2(r+4)^2\cdots(r+2n)^2}[/math] with respect to r, in order to show that [math]c'_{2n}(0)=\frac{(-1)^{n+1}H_n}{2^{2n}(n!)^2}[/math]

My solution involved expanding the denominator of [imath]c_{2n}(r)[/imath] into a polynomial of degree n squared with undetermined coefficients b

_{n}except for b_{1}and b_{0}, since I was only interested in the case when r=0, and thus all the other terms wouldn't matter. Differentiating that via use of the chain rule and such did give me the correct result. However, the professors solution says this: [math]c'_{2n}(r)=(-1)^n\left(\frac{-2}{(r+2)^3(r+4)^2\cdots(r+2n)^2} -\frac{2}{(r+2)^2(r+4)^3(r+6)^2\cdots(r+2n)^2} -\cdots-\frac{2}{(r+2)^2(r+4)^2\cdots(r+2n)^3}\right)[/math][math]=\frac{2(-1)^{n+1}}{(r+2)^2(r+4)^2\cdots(r+2n)^2}\left(\frac{1}{r+2}+\frac{1}{r+4}+\cdots+\frac{1}{r+2n}\right)[/math]While that gives the right result for r=0, it strikes me as if that differentiation is neglecting the product rule, and thus shouldn't work. Am I missing something that does make it work out as he did it, or is his solution flawed?

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