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This question came up between me and my teacher when we discussed whether the following is true or not.

Given that $\lim_{x \to a} f(x) = 0,$ is it true that $\lim_{x \to a} \frac{\ln(1+f(x))}{f(x)} = 1?$

It is widely known (e.g., by L'Hopital's Rule) that

$$\lim_{x \to 0} \frac{\ln(1+x)} x = 1,$$

but we wanted to know if it could be generalized.

Now, there are some trivial counterexamples like $f(x)=0$ or any function where it’s flat around $0,$ but my teacher argued that it could be generalized.

So, I used $f(x)=\sin(1/x)x^2$ as a counterexample. I argued that this diverges since it is goes to $1$ for most of the time, but there are values that are undefined, so it is divergent. But my teacher argues that it does indeed converge to $1.$

Can anyone solve this for us? Thanks.

Dylan C. Beck
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AntDochi
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  • This is not a question about Mathematica software? – flinty Jun 01 '20 at 12:15
  • The limit indeed converges to 1: https://www.wolframalpha.com/input/?i=Limit+ln%281%2Bx%5E2+sin%281%2Fx%29%29%2F%28x%5E2+sin%281%2Fx%29%29+x-%3E0 – yarchik Jun 01 '20 at 12:20
  • Could you explain why? My teacher also showed me the wolframalpha result but I don’t understand why it is convergent. – AntDochi Jun 01 '20 at 14:03
  • Relevant: https://math.stackexchange.com/questions/919798/the-limit-of-composition-of-two-functions and https://math.stackexchange.com/questions/2831604/limit-of-composite-functions and https://math.stackexchange.com/questions/1726806/limit-of-the-composition-of-two-functions-with-f-not-necessarily-being-continuou and https://math.stackexchange.com/questions/1069642/finding-a-limit-using-change-of-variable-how-come-it-works and https://math.stackexchange.com/questions/167926/formal-basis-for-variable-substitution-in-limits – yarchik Jun 01 '20 at 15:02
  • You need additional hypothesis that $f(x) \neq 0$ as $x\to a$. However in the specific case when $f(x) =x^2\sin(1/x)$ we can say that the limit is $1$ by using a more relaxed definition of limit. Ultimately questions like these are more a matter of convention than mathematical argument. – Paramanand Singh Jun 01 '20 at 17:11
  • So I proposed that the conjecture hold when there exists a > 0 that for all x 0<|x-a|< f(x) is not 0. However I guess it can be seen as a matter of how limits are defined and calculated. If anyone else could add to this or prove any of us wrong, please do! – AntDochi Jun 01 '20 at 22:34
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    Might we have $\ln(1+f(x)) = f(x) - f(x)^2 +f(x)^3 - \ldots = f(x) (1+O(f(x))$, so that $\ln(1+f(x))/f(x) = 1+ O(f(x))\to 1$? If $f(x_0)=0$ for some $x_0$ near $0$, just define the quotient to be $1$... – Integrand Jun 04 '20 at 20:28
  • Well, I think this comes down to convention and how we define limits in the first place, as Paramanand said previously. Since, when you’re cancelling out the f(x), it is being assumed that f(x) is not zero. Of course, by a looser definition of the limit it could be argued that this still holds, but I don’t think it fully answers our question about this exact problem when f(x)=0. Interesting take though! – AntDochi Jun 05 '20 at 12:10

1 Answers1

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We will assume that $\lim_{x \to a} f(x) = 0.$ Let us investigate the quantity$$\lim_{x \to a} \frac{\ln(1 + f(x))}{f(x)}.$$ (1.) Given that $f(x)$ is differentiable on some open interval $(b, c)$ containing $a$ (except possibly at $x = a$) and $f'(x) \neq 0$ for all $b < x < c$ (except possibly at $x = a$), then by L'Hopital's Rule, $$\lim_{x \to a} \frac{\ln(1 + f(x))}{f(x)} = \lim_{x \to a} \frac{f'(x)}{(1 + f(x)) f'(x)} = \lim_{x \to a} \frac{1}{1 + f(x)} = 1.$$

Unfortunately, if $f(x)$ has infinitely many roots in a neighborhood of $a,$ then we cannot use this argument. Particularly, for the function $f(x) = x^2 \sin \bigl(\frac 1 x \bigr)$ and $a = 0,$ one can prove that $f'(x) = 2x \sin \bigl(\frac 1 x \bigr) - \cos \bigl(\frac 1 x \bigr)$ has infinitely many roots in a neighborhood of $a = 0.$

(2.) Considering that $-1 < f(x) \leq 1$ in a neighborhood of $a,$ we may use the power series expansion of $\ln(1 + x)$ to obtain the formal power series $$\ln(1 + f(x)) = f(x) - \frac{[f(x)]^2}{2} + \frac{[f(x)]^3}{3} - \frac{[f(x)]^4}{4} + \cdots,$$ from which it follows immediately that we have $$\frac{\ln(1 + f(x))}{f(x)} = 1 - \frac{f(x)}{2} + \frac{[f(x)]^2}{3} - \frac{[f(x)]^3}{4} + \cdots.$$ By hypothesis that $\lim_{x \to a} f(x) = 0,$ we conclude $$\lim_{x \to a} \frac{\ln(1 + f(x))}{f(x)} = \lim_{x \to a} \biggl(1 - \frac{f(x)}{2} + \frac{[f(x)]^2}{3} - \frac{[f(x)]^3}{4} + \cdots \biggr) = 1.$$

Dylan C. Beck
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    I guess my question to that would be could you even set an interval (b,c) containing 0 that f’(x) isn’t 0 since for any value for b or c, you could prove that there is a f’(x)=0 point small than that. Also how should I analize L’Hopital’s rule when the expression is undefined at certain values of x? – AntDochi Jun 01 '20 at 22:08
  • For one, why are you concerned about the interval containing $0?$ I tried to be quite general with my answer, assuming only that $f(x)$ is differentiable. (I don't think that's asking too much.) For L'Hopital's Rule, I made sure that my $f(x)$ satisfies all of the necessary requirements; however, if some function you care about has a removable discontinuity at a point, just take $f(x)$ to be its continuous extension. – Dylan C. Beck Jun 02 '20 at 12:27
  • I’m sorry if my previous comment seemed aggressive in some way, I wan’t trying to discredit your answer or criticize your take. I was just curious about this specific instance when x->0 since my opinion is that L’Hopital’s rule can’t be utilized here. Of course we could simply substitute f(x) for a continuous counterpart but I think it would defeat the purpose of me and my teacher’s question. We agree that it converges to 1 if the discontinuous parts are ignored but we are wondering how to analyze the discontinous parts. – AntDochi Jun 03 '20 at 23:16
  • In the proof, $a$ is arbitrary, so even if $a = 0,$ everything I have written still holds. I guess I just don't understand your concern. – Dylan C. Beck Jun 04 '20 at 04:53
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    well when a=0 there would not exist an open interval (b,c) that satisfies ′()≠0 right? since there are infinite points where it is zero around x=0. That’s what I would take from L’Hopital’s rule. – AntDochi Jun 04 '20 at 12:08
  • I see that you have edited your solution based on our conversation! I like how you rephrased my question about my sepcific function f(x), I think it is more clear than what I initially proposed and I thank you for contributing to my inquiry. So, here’s my take on this new approach. With the power series expansion of ln(1+f(x)) and the subsequent cancelling out of f(x), that process is based on the assumption that f(x) ≠ 0. So, I think that my proposal still stands that the limit does not equal 1. However, I am enjoying this little debate we’re having and hope to hear any of your new ideas! – AntDochi Jun 05 '20 at 12:23
  • Well, if $f(x) \equiv 0,$ then $\ln(1 + f(x)) \equiv \ln(1) = 0,$ in which case the object for which you would like to compute the limit as $x \to a$ is $0/0,$ and this could be anything. So, in this case, I fear that the question is not well-defined. – Dylan C. Beck Jun 06 '20 at 04:34
  • Yes limits for 0/0 can be anything. However, in the case of ln(1+x)/x, it is in the form of 0/0 but the limit does converge to 1. In fact in most cases, the limit of ln(1+f(x))/f(x) does indeed converge to 1. The core question I would like help with is whether or not this holds true or all f(x). I have previously metioned that for some functions like f(x)=0, it is trivial that the limit is undefined. However, I wanted to know how to approach this question when f(x)=sin(1/x)x^2 since I am unsure if my claim that the limit is undefined is mathematically rigorous. – AntDochi Jun 06 '20 at 09:42
  • It's not. As you can see, for $f(x) = x^2 \sin \bigl(\frac 1 x \bigr),$ we have that $-1 < f(x) \leq 1$ in a neighborhood of $a = 0$ (e.g., $-1 < x < 1$), hence by case (2.) above, the limit you wish to evaluate is 1. – Dylan C. Beck Jun 06 '20 at 18:22
  • Sorry, I’m having trouble understanding your argument. You said previously that the question is not well defined but still state that the limit does indeed converge to 1. For case (2), it does not hold true since your act of cancelling out f(x) requires a condition that f(x)≠0. However, as x>0, f(x) is zero infinitely many times. Therefore the cancelling out cannot be done before the limit is taken. I understand that at f(x)=0, the limit is undefined. However, that is the essense of what we’re trying to solve. I believe just ignoring the discontinous points are ignoring the problem. – AntDochi Jun 07 '20 at 01:04
  • Even though $f(x)$ is zero infinitely many times in a neighborhood of 0, the argument in (2.) still holds because $f(x) = x^2 \sin \bigl(\frac 1 x \bigr)$ is not identically zero. If the function you are dealing with is identically zero, then of course, we cannot even consider the quotient $\frac{\ln(1 + f(x))}{f(x)}.$ – Dylan C. Beck Jun 07 '20 at 02:17
  • Then how about considering the function f(x) = [x] when x>0 and f(x)=x when x<0 where [x] means the largest integer smaller than or equal to x. The function woulnd’t be identically zero, but would be flat on the right side of x=0. What if that flat piece of the function was infinitely small around x=0? Also, it still holds that you would need a condition that f(x)≠0 for the cancelling out to be rigorous, or at least change the domain so that f(x)≠0, in which case the limit is indeed 1. However, I don’t think that argument holds when you consider that there are repeated undefined points as x>0. – AntDochi Jun 07 '20 at 03:10
  • In this case, the limit as $x \to 0^+$ is not well-defined because $f(x) = 0$ for $0 \leq x < 1,$ so the limit as $x \to 0$ is not well-defined. – Dylan C. Beck Jun 07 '20 at 03:39
  • I see! I think that argument could be said for the initial question. As the limit is repeatedly not defined as x>0, wouldn’t the limit ln(1+f(x))/f(x) also be undefined and diverge instead of converging to 1? – AntDochi Jun 07 '20 at 03:45
  • No. Considering that $f(x) = x^2 \sin \bigl(\frac 1 x \bigr)$ is not identically zero or $-1,$ the quotient $\frac{\ln(1 + f(x))}{f(x)}$ makes sense; its domain is all $x$ such that $f(x) \neq 0$ and $f(x) > -1.$ – Dylan C. Beck Jun 07 '20 at 17:59