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I need to find: $$\lim_{x \to \infty} \frac{\ln(x+1)}{x}$$ without using L'Hospital, integrals or derivative!

I would like your help in order to know i'm not doing something invalid.

I was thinking of the following way:

$$ \lim_{x \to \infty } \frac{\ln(x+1)}{x} = \lim_{x \to \infty } \frac{1}{x} \ln(x+1) = \lim_{x \to \infty } \ln(x+1)^\frac{1}{x} $$

and because $\frac{1}{x}$ tend to $0$ as $x$ approaches $\infty$, than $(x+1)^\frac{1}{x}$ tends to $1$ and $\ln(1) = 0$ and the limit is $0$.

But this way sounds off. I was trying also to express $x$ with $t$ and use $e$ (euler) somehow, but it didn't work.

Is my way valid? If not, do you have a better way without using L'Hospital's rule?

Thanks!

Edit:

While reading answers I was thinking of another way,

$$\lim_{x \to \infty} \frac{\ln(x+1)}{x} = \lim_{x \to \infty} \frac{\ln(x(1+\frac{1}{x}))}{x} = \lim_{x \to \infty} \frac{\ln(x)}{x} + \frac{\ln(1+\frac{1}{x})}{x} = $$ $$\lim_{x \to \infty} \ln(\sqrt[x]{x}) + \frac{\ln(1+\frac{1}{x})}{x}$$

now $\sqrt[x]{x}$ tends to $1$ as x approaches to $\infty$, hence $\ln(\sqrt[x]{x})$ tends to $0$ (is it valid to say so?)

and $\frac{\ln(1+\frac{1}{x})}{x}$ tends to $0$ is pretty obvious.

Am on missing something in this one?

Yeregb
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  • I think showing the limit of $(x+1)^{1/x}$ is not straightforward. You should prove in more descent way. You said $1/x$ tends to zero but on the other hand $x+1$ goes to infinity. – Red Phoenix Dec 10 '20 at 20:43
  • About your edit, how do you prove that $\sqrt[x]{x}$ tends to $1$ as $x\rightarrow +\infty$ ? You do it with the use of $\lim\limits_{x\rightarrow +\infty}\frac{\log x}{x}=0$ since $\sqrt[x]{x}=e^{\frac{\log x}{x}}$ (as far as I know). Therefore your proof is circular and thus is not valid. – Tuvasbien Dec 10 '20 at 23:17
  • @Yeregb Hi! It's been a while. I hope you're staying safe and healthy during the pandemic. I've reached out to contact you a few times, but am unsure whether you've received the notes? If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) – Mark Viola Mar 22 '21 at 18:13

5 Answers5

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Without L'Hospital Rule (or series for that matter!) you are left with using some basic properties of functions. For example set $\ln(x+1)=t$ so that $x=e^t-1$ and now you have an expression $\frac{t}{e^t-1}$ with $t $ going to infinity. To what extend does your teacher want you to prove that an exponential functions grows much faster than a linear term?

amWhy
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imranfat
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  • I'm not sure... is it something that is usually obvious when using it in a solution or it requires some sort of proof? – Yeregb Dec 10 '20 at 20:43
  • It's not uncommon that L'Hospital'r Rule or any other derivative product cannot be used to find a limit. But typically some "rules of the game" are then given. In Precalculus we learn that an exponential function grows faster than a polynomial function. It's not very formal in terms of "proof" and from a strict analytical standpoint, barely a proof. It wasn't clear to me. Several answers have been provided to you, some with more rigor than others. Pick your choice that suits the best for your course – imranfat Dec 10 '20 at 20:51
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Let $c\in\mathbb{R}$ and $f(x)=\log(x)-\sqrt{x}+c$, then $f'(x)=\frac{1}{x}-\frac{1}{2\sqrt{x}}=\frac{2-\sqrt{x}}{2x}$, therefore the maximum of $f$ is $f(4)=\log 4-2+c$. We take $c=2-\log 4$, we thus have $f(x)\leqslant 0$ for all $x>0$, which leads to $\frac{\log(x)}{x}\leqslant\frac{1}{\sqrt{x}}+\frac{\log 4-2}{x}\underset{x\rightarrow +\infty}{\longrightarrow}0$.

Tuvasbien
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I will use the substitution $y=\ln(1+x)$. Let $n$ be $n=[y]$ Then $x=e^y-1$. We can simply estimate for $x>10$ (say): $$ 0\le \frac{\ln(x+1)}x=\frac y{e^y-1}<\frac y{e^y/2}<\frac y{2^y/2} < \frac {n+1}{2^{n-1}} = \frac {n+1}{(1+1)^{n-1}} < \frac {n+1}{\binom{n-1}2} <\frac {2n}{\frac n2\cdot\frac n2\cdot \frac 12} =\frac {16}n\to 0 \ . $$

dan_fulea
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  • Of course, $e^y-1>\frac12 y^2$ for $y> 0$. And you could have simply written $$\frac{\log(x+1)}{x}=\frac{y}{e^y-1}<\frac{2}{y}$$and you're done. ;-) – Mark Viola Dec 10 '20 at 21:03
  • @MarkViola, I think dan only wants to use the inequality $e\gt2$, not an inequality that requires its own separate proof. – Barry Cipra Dec 10 '20 at 21:15
  • @BarryCipra Well, how does one know that $e>2$ without a definition and subsequent proof? – Mark Viola Dec 10 '20 at 21:47
  • @MarkViola What is your definition of $e$? Is maybe one of $e=1/0!+1/1!+\dots$ or the convergence of the increasing sequence $(1+1/n)^n$ starting with $2$, $2.25$, $2.37037...$ ?! – dan_fulea Dec 10 '20 at 21:58
  • @dan_fulea Yes, I know Dan. I was responding to Barry's comment. In fact, there are many equivalent characterizations of the exponential function $e^x$ from which $e^1$ can be evaluated. And from any of those characterizations, it is also straightforward to show that $e^y-1>\frac12 y^2$. That was the point of my comment. Is there something confusing herein? – Mark Viola Dec 10 '20 at 22:18
  • @MarkViola, you are quite right, even $e\gt2$ requires definition and proof. But I think the proof is a step down from what's required for the inequality in your comment, at least for the $(1+y/n)^n\to e^y$ definition of $e^y$. – Barry Cipra Dec 11 '20 at 00:05
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It is easy to show (See Here) that $\log(x)<\sqrt{x}$ for all $x>0$. Therefore, we have

$$\frac{\log(1+x)}{x}\le \frac{\sqrt{x+1}}x$$

And you can finish now.

Mark Viola
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    "I would like your help in order to know i'm not doing something invalid." I don't see your answer to that question, asked first. – amWhy Dec 10 '20 at 20:32
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    @amWhy Well, the last question is "Do you have a better way without using L'hospital rule?" So, it answers that question. Moreover, the lead statement is "I need to find ... without using L'Hospital's or Integrals." – Mark Viola Dec 10 '20 at 20:34
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    Is it easy to show $\ln(x)<\sqrt{x}$ though? Also, this immediately gives $\frac{\ln(x+1)}{x}<\frac{\sqrt{x+1}}{x}$, which streamlines it a bit. – pancini Dec 10 '20 at 20:39
  • @ElliotG Indeed. – Mark Viola Dec 10 '20 at 20:41
  • Please show $\log x<\sqrt x$ for all $x>0$. (I don't know why we need this on $(0,1)$, but you claim so. Then please show the other simple thing, $\log(1+y)\le y$. (For that unspecified $y=1/x$.) Best without l'Hospital or integrals. (Why not $\log(1+x)\le \log(2x)$ for $x\ge 1$ for an easy life.) It is not my downvote, but think about this policy of quickly typing an answer. – dan_fulea Dec 10 '20 at 20:47
  • @dan_fulea I've provided a link. Does that suffice? – Mark Viola Dec 10 '20 at 20:49
  • Ok, then you can provide directly a link for $\ln(x+1)/x\to 0$ instead of the link. And now the link shows me something i do now want for an integer $k$. You come closer to a downvote. – dan_fulea Dec 10 '20 at 20:54
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    @dan_fulea If you have a closer read at the answer I posted in the link, the inequality derived applies to all $x\in \mathbb{R}$, not just integer values. For that specific post, the OP was analyzing a series. That was the primary motivation. But nowhere in the development was $x$ restricted to be an integer. Does that help better? – Mark Viola Dec 10 '20 at 20:55
  • @MarkViola If, and this is a big if. Since we are showing something very simple using (first less polished than now) a more complicated inequality, which was edited, and as it came it came without any link. Now it comes with a link, and in the link there is an integer $k$. Then the comment tells me to look into the proof, the proof uses an other link namely this one :: https://math.stackexchange.com/questions/1589429/how-to-prove-that-logxx-when-x1/1590263#1590263 with a questionable definition of $e^x$, namely the one as $\lim$ of $(1+x/n)^n$. Yes, it is a definition. – dan_fulea Dec 10 '20 at 22:07
  • @dan_fulea You seem to be confused about things Dan. Let's part here as friends as your comments clearly show a lack of purpose. I used a well-known inequality and provided a link to a methodology which proves that inequality. If you believe that one should repeat a development every time one appeals to a theorem or previously derived result, then you are in strong disagreement with the research community. – Mark Viola Dec 10 '20 at 22:21
  • @MarkViola We will never part, but just stay friends - comments can never change the fact that we are all on the same side of science. – dan_fulea Dec 11 '20 at 01:53
  • @dan_fulea I'm happy to call you my MSE friend. Stay safe and healthy. – Mark Viola Dec 11 '20 at 03:12
  • @Yeregb Hi! It's been a while. I hope you're staying safe and healthy during the pandemic. I've reached out to contact you a few times, but am unsure whether you've received the notes? If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) – Mark Viola Mar 22 '21 at 18:13
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$\ln\nearrow\, $ thus for $\, x\gg 1\, $ then $\, x+1<2x\implies 0<\dfrac{\ln(x+1)}{x}<\dfrac{\ln(2x)}{x}=\underbrace{\dfrac{\ln(2)}{x}}_{\to 0}+\dfrac{\ln(x)}{x}$

So we have reduced the problem to studying $f(x)=\dfrac{\ln(x)}{x}$

$f'(x)=\dfrac{1-\ln(x)}{x^2}<0$ at infinity, so $f\searrow$ at infinity and in particular this means that $f$ is bounded by some $M$ at infinity.

Now we can use the multiplicative to additive property of logarithm,

$f(x)=\dfrac{2\ln(\sqrt{x})}{(\sqrt{x})^2}<\dfrac{2M}{\sqrt{x}}\to 0$

zwim
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