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Consider $\lim\limits_{x \to +\infty} \frac{f(x+1)}{f(x)}$

Intuitively, it seems to me that this limit should equal to 1 because "infinity" and "infinity+1" is essentially the same thing.

However, I'm not sure if this is completely correct.

Secondly, if it is correct, I have no idea how to show this using the limit laws I have learned (or even where to begin since I don't know that either limit exists seperately so I can't just seperate them)

Masacroso
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Milk Man
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  • Can you show what this would mean for an example function, say, $f(x) = x^2$? Please add to your post how this question would apply to this f(x). – amWhy May 19 '18 at 19:46
  • it depends on your $f(x)$ is. If, for example , $f(x)=x$ then $\lim_{x\to\infty}\frac{f(x+1)}{f(x)} = 1$ – The Integrator May 19 '18 at 19:47
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    There is no universal answer. Consider such examples as $x$, $e^x$ and $e^{x^2}$. – Michal Adamaszek May 19 '18 at 19:47
  • You can experiment your intuition with some simple function as $e^x$. What is the result in this case? – Emilio Novati May 19 '18 at 19:48
  • Were you given a function $f(x)$ to work with, Milk Man? – amWhy May 19 '18 at 19:48
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    You should at least specify explicitly the behaviour of $f$ at infinity (you only suggests it in your first sentence). – zwim May 19 '18 at 19:49
  • The general case seems to have been answered very well by now. Thank you to those who answered. One specific case I'm still unsure of is when lim as x--> infinity of f(x) =0 (because if I try using the limit laws, I get 0/0) – Milk Man May 20 '18 at 02:50

3 Answers3

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Even under the assumption that $\lim_{x\to\infty} f(x) = \infty$ (implicit in your question), we cannot say much.

  • The limit can be $1$: take $f(x) = x$: $$\lim_{x\to\infty} \frac{x+1}{x} = 1$$

  • The limit can be any positive number $a>1$: take $f(x) = a^x$: $$\lim_{x\to\infty} \frac{a^{x+1}}{a^x} = a$$

  • The limit can be $\infty$: take $f(x) = e^{e^x}$: $$\lim_{x\to\infty} \frac{e^{e^{x+1}}}{e^{e^x}} = \infty$$

  • The limit can fail to exist: take $f(x) = \begin{cases} x &\text{ if } x\in 2\mathbb{N}\\2x &\text{ if } x\in 2\mathbb{N}+1\\ x &\text{otherwise.}\\\end{cases}$: then$$\lim_{n\to\infty }\frac{f(2n+1)}{f(2n)} = 2,\qquad \lim_{n\to\infty }\frac{f(n\pi+1)}{f(n\pi)} = 1$$

(to have the same "limit of the ratio fails to exist" behavior for a continuous function $f$, one can also consider $f(x) = x(2+\cos x)$: $\liminf_{x\to\infty} \frac{f(x+1)}{f(x)} = \frac{1}{2}$, $\limsup_{x\to\infty} \frac{f(x+1)}{f(x)} \simeq 1.73 > 1$, and indeed $\lim_{x\to\infty} f(x) = \infty$.)

Note: the ratio can have a limit $1\leq \ell\leq +\infty$, or no limit; in the latter case, it can even have a $\liminf$ equal to $0$. However, it cannot have a limit $\ell < 1$, as otherwise the function $f$ must be decreasing for $x$ big enough and then cannot satisfy $\lim_{\infty} f = \infty$.

Clement C.
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    Answer left as community wiki, feel free to edit/improve. – Clement C. May 19 '18 at 19:59
  • Nice work, @ClementC. – amWhy May 19 '18 at 20:14
  • @amWhy Thanks! By now, I think what's in this answer should cover all the cases for $f$ such that $f\to\infty$; except maybe "limit fails to exist but $f$ is continuous" (the function in the fourth bullet is not continuous). – Clement C. May 19 '18 at 20:16
  • I can think of $f(x) = x(2+\cos x)$ for that. – Clement C. May 19 '18 at 20:18
  • @ClementC. You can steal my example for $\sin x$ :P – user May 19 '18 at 20:24
  • @gimusi I focused on $f$ such that $f\to\infty$ :) (I believe that's what the OP had in mind, and also I didn't want to have a function taking value $0$ in the denominator). But again, it's Community Wiki (no one gets points, everyone can contribute) – Clement C. May 19 '18 at 20:25
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    @ClementC. The OP doesn't make any claim about the fact that $f(x)\to \infty$. His/Her point of counfusion is that $\infty+1\approx\infty \implies f(x+1)\approx f(x)$. – user May 19 '18 at 20:28
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Let consider as counterexample $f(x)=e^x$ then

$$\frac{f(x+1)}{f(x)}=\frac{e^{x+1}}{e^x}=e$$

or also

$$f(x)=\sin (x)\implies \frac{f(x+1)}{f(x)}=\frac{\sin(x+1)}{\sin x}\to N.E.$$

Thus in general the limit depends upon the particular function considered.

user
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    It depends on the function; why showcase only $f(x) = e^x$? As stated, the question is about function $f(x),$ in general, and seeking $\lim_{x\to \infty} \frac{f(x+1)}{f(x)}$. You choose to showcase only one function? – amWhy May 19 '18 at 19:50
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    please to not steal others people's arguments so openly... Elaborate on the subject instead and make a fully developped answer. – zwim May 19 '18 at 19:50
  • since the OP is claiming that th elimit should be always equal to 1 – user May 19 '18 at 19:50
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    @zwim how is this stealing? no one holds a patent on the function $e^x$ – The Integrator May 19 '18 at 19:51
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    @zwim steal? do you mean $e^x$ is copyrighted? – user May 19 '18 at 19:52
  • Now you just stole The integrator's comment, gimusi! Think for yourself. – amWhy May 19 '18 at 19:53
  • @amWhy That's indeed a counterexample since it is $\neq 1$ – user May 19 '18 at 19:53
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    I really didn't stole anything, maybe someone else gave the same example in a comment while I was writing here, it doesn't seem a big idea to me. It is just a trivial counterexample. – user May 19 '18 at 19:55
  • My point, @gimusi (and no I did not downvote), is that yes, with some functions, the limit may be $1$, but not so with others. I.e., as I said in my comment here: it depends on what $f(x)$ is. – amWhy May 19 '18 at 19:55
  • agreed, no one holds a monopoly on counter examples. Just because people say the same answer doesn't mean its wrong, if the answer is correct its bound to be the same . – The Integrator May 19 '18 at 19:56
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    why the downvote? This is a perfect counter-example to the question, that is, it shows that its not true in general that $\lim_{x\to\infty}f(x+1)/f(x)=1$ – Masacroso May 19 '18 at 19:56
  • @amWhy you are not wrong , maybe you could also add some more counterexamples in this post. teamwork – The Integrator May 19 '18 at 19:57
  • @TheIntegrator There is now an answer which is much more thorough. – amWhy May 19 '18 at 20:00
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    @amWhy Many times I've seen counterexamples given and accepted as an answer. If the claim is "it seems to me that this limit should equal to 1" a counterexample indeed suffices to disprove the claim. Of course one cannot like this kind of answer and can add his own with more details. But I'm sill convinced that mine is a proper answer to the OP. Thanks anyway for your suggestions. – user May 19 '18 at 20:09
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    The point is is to be helpful to the OP and to future readers. Your answer only shows that the limit need not be $1$, but could be $e$, or $a$ if $f(x) = a^{x}$. The other answer is much more thorough and explanatory. The point of answering is not just to get upvotes or an accepted answer. It is to be as helpful as one can. Anything you add at this point (like using $f(x) = \sin x$ will be seen by me as simply an attempt to gain more favor, as anything you add has already been said on this post. – amWhy May 19 '18 at 20:13
  • @amWhy Yes I undestand your point, indeed now I've updated adding another example and explaining that "in general the limit depends upon the particular function considered". – user May 19 '18 at 20:15
  • @amWhy I've just read now your comment in which you accuse me to have stolen the comment by integrator. Are you joking or you are trying to get me angry? I really hope you are joking! – user May 19 '18 at 20:27
  • Joking... sorry if it didn't come across as a joke. I know humor can sometimes not translate well across languages. I should have thought of that when I was joking. It was joking in reference to the comment about you "stealing" an example, a comment which I do not agree with. – amWhy May 19 '18 at 20:30
  • @amWhy Ah ok, indeed I wasn't sure about that. Thanks for the clarification! – user May 19 '18 at 20:31
  • @gimusi for completeness you could also include the case $\lim_{x\rightarrow \infty} f(x)$ exists. – clark May 19 '18 at 22:26
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Consider a random series taking the valus of $1$ or $2$ with same probability: $r:\,\mathbb{N}^{+}\to\left\{ 1,2\right\}$ where $\lim_{N\to\infty}\frac{1}{N}\sum_{i=1}^{N}r\left(i\right)=\frac{3}{2}$. The possible quotients $\frac{r\left(i\right)}{r\left(i+1\right)}$ then take random values in $\left\{ \frac{1}{2},1,2\right\}$ with probabilities $\frac{1}{4}$, $\frac{1}{2}$ and $\frac{1}{4}$ respectively. There is no limit at all!

If however $\lim_{x\to\infty}f\left(x\right)$ exists, what can we say about the quotient then?

Consider the series $h:\,\mathbb{N}^{+}\to\mathbb{Q}$ with $h\left(i\right)=\frac{1}{2}^{i}$. Then $\frac{h\left(i\right)}{h\left(i+1\right)}=2$.

So we have a limit, but it is not $1$...

But there need not be a limit of the quotient even in the case that $f\left(x\right)$ approaches a limit:

With our random function $r$ as defined above, define a series $d:\,\mathbb{N}^{+}\to\mathbb{Q}$ with $d\left(1\right)\,:=\,1$ and $d\left(i+1\right)=\frac{d\left(i\right)}{r\left(i\right)}$. The limit is $\lim_{i\to\infty}d\left(i\right)=0$. But $\frac{d\left(i\right)}{d\left(i+1\right)}=r\left(i\right)$ takes no limit.

By continuos extension of the above series it is obvious there are similar examples for continuos functions. Use splines to construct examples for differentiable functions.

Bottomline: Without some additional knowledge about the function, it cannot even be guaranteed that the quotient takes any limit, let alone $1$.

mol3574710n0fN074710n
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