2

Find $\lim_{x \to \infty} (\frac{1}{e} - \frac{x}{x+1})^{x}$

This is of the form $(\frac{1}{e} - \frac{1}{(1/∞) + 1})^{∞}$

Now I'm stuck..

How should I go further. If I find the limit of the function inside the bracket, it is $(1/e - 1)$, which is a negative number less than $1$. It is approx $-0.632$. So its power $(=∞)$ should make the number $0$.

But it is given in answer that the limit doesn't exist.

I think I'm missing some basic concept.

Any suggestion$?$

Mathaddict
  • 2,300
  • The limit exists, and it is zero. – Dr. Sonnhard Graubner Dec 02 '19 at 15:15
  • It is given in answer that $(-$ ve value$)^{∞}$ doesn't exist. – Mathaddict Dec 02 '19 at 15:16
  • @Dr.SonnhardGraubner : This limit can't exist because the base is negative for sufficiently large $x$, and so the expression is undefined for arbitrarily large $x$ (specifically: $(\textrm{negative irrational number})^{n+\frac12}$ is undefined). – MPW Dec 02 '19 at 15:41
  • @MPW: it is not undefined; it simply requires a branch cut to be defined. – robjohn Dec 02 '19 at 16:32
  • robjohn : Yes, perhaps I can soften this claim. I have addressed the consideration of complex values in my answer and agree with @Dr.SonnhardGraubner that in one sense the limit can be considered as existing and equaling zero. However, in a real-variable context, I would say the limit does not exist. – MPW Dec 02 '19 at 16:42
  • @MPW You are completely wrong, the limit exists and it is equal to zero. Please refer to the discussion here and please consider to remove your comments from my answer which is really misleading in particular for the asker. – user Dec 02 '19 at 20:13
  • @robjohn Sorry to disturbing you. If possible, I would appreciate to have also your prominent advice on the correctness or fallacy of the answer I've given. Thanks – user Dec 02 '19 at 20:44

3 Answers3

3

Note that $$\frac1e-\frac{x}{x+1} = \frac1e-\frac{x+1-1}{x+1} = \frac1e-1+\frac{1}{x+1}$$ $$= \underbrace{-\left(1-\frac1e \right)}_{\textrm{negative constant}} + \underbrace{\frac{1}{x+1}}_{\textrm{tends to zero}}$$

so the expression $\left(\frac1e-\frac{x}{x+1}\right)^x$ is undefined for some arbitrarily large values of $x$ (for example, if $x=n+\frac12$ for large integers $n$).

This means the limit can't exist. The reason is that if the limit did exist, then it would would also equal $$\lim\limits_{n\to\infty}\left(\frac1e-\frac{a_n}{a_n+1}\right)^{a_n}$$ for sequence $(a_n)$ which increases without bound. But we have demonstrated such a sequence for which every one of the terms is undefined.

Addendum: I point out that notation is critical here. In writing "$x\to\infty$", the implication is that $x$ grows without bound taking on all real values in the process.

However, if you used the notation "$n\to\infty$" instead (and used the variable $n$ instead of $x$), convention is that you have a sequence and $n$ grows without bound taking on integer values only.

This is purely conventional usage and is not strictly well-defined, but it is certainly largely accepted and used. Context would clarify, and in this case it is critical.

Addendum 2: If you allow complex values of the expression and multiple-valuedness of them, then the limit does indeed exist and equals zero. Here's why. We may consider only $x$ sufficiently large so that the base is always negative (as described above). Then we can write $$\left(\frac1e-\frac{x}{x+1}\right)^x = \left(r(x)e^{\pi i}\right)^x$$ where $r(x) = 1-\frac1e - \frac1{x+1}$ is strictly positive. This expression is multiple-valued and takes values (for integral $k$) $$r(x)^x\cdot e^{(2k+1)\pi xi} = r(x)^x\cdot e^{i\theta_k(x)}$$ where $\theta_k(x)=(2k+1)\pi x$ for integral $k$.

As others have shown, $r(x)^x\to 0$, so the value of $\theta_k(x)$ doesn't matter; the angles grow without bound, but the magnitude shrinks to zero, so in the end the value spirals to $0$ as $x\to\infty$.

This is true for all of the multiple values (the values corresponding to fixed $k$), so the limit exists independent of the choice of $k$. The expressions are well-defined (multiple-valuedly) for all $x$, and all tend to zero. So the limit is zero.

MPW
  • 43,638
  • 1
    I think that we can assume that the limit is taken for $x \to \infty$ such that the expression id well defined, in this case the limit exists and it is equal to zero. – user Dec 02 '19 at 15:53
  • Your argument for complex is fine but the limit exists also on reals if we refer to the more general definition of limit. Of course it is mainly a matter of definition, but I think that we should prefer those definitions which promote the existence for the limit since they are more general. More details are given in the link attached to my answer. – user Dec 02 '19 at 16:59
  • Can you please explain why the expression $(\frac{1}{e}− \frac{x}{x+1})^x$ is undefined for some arbitrarily large values of x. why $(negative;; irrational;; number)^ {n+ \frac{1}{2}}$ is undefined ? @MPW – ankit Dec 02 '19 at 17:53
  • Your answer is unnecessarly complicated. Refer to the link I've provided for a general discussion of the issue. According to the definition of limit, the limit exists and it is equal to zero, also for x real. We don't need nothing else. If you are reffering to high school definition, the revise the more general definition. – user Dec 02 '19 at 20:16
  • 1
    @ankit : It doesn't need to be "negative irrational number". Just any negative number raised to the power $n+1/2$ is undefined (meaning it isn't a real number). This is assuming you are dealing with real-valued functions of a real variable. – MPW Dec 02 '19 at 20:57
  • Thanks a lot @MPW. Now, I understood, it is like $(-1)^{1/2} = i $ which is not a a real. So, anything multiplied by it, will also be not real, that's why $(any;; negative;; number)^{n+\frac{1}{2}}$ is not real. – ankit Dec 03 '19 at 05:29
2

If your limit converges to $L$ then $$ \ln L = \ln \left(\lim_{x \to \infty} \left(\frac{1}{e} - \frac{x}{x+1}\right)^x \right) = \lim_{x \to \infty} \ln \left(\left(\frac{1}{e} - \frac{x}{x+1}\right)^x \right) $$ and now exponent $x$ can be pulled outside the log...

UPDATE

Note that $1/e < 1$ so for $x$ large enough the denominator is negative. Hence, if $x \in \mathbb{R}$, the limit does not exist. If, on the other hand, $x \in \mathbb{N}$, this is well-defined and the calculations will make sense...

gt6989b
  • 54,422
  • 1
    Some negative value inside $ln$ is not defined, so limit is not defined. – Mathaddict Dec 02 '19 at 15:19
  • Final answer is indeed $0$. – Harshal Gajjar Dec 02 '19 at 15:20
  • @Harshal Gajjar how? – Mathaddict Dec 02 '19 at 15:23
  • 2
    This is incorrect. The limit does not exist because the expression becomes undefined for arbitrarily large $x$. – MPW Dec 02 '19 at 15:37
  • 2
    It depends on whether $x\in\mathbb{Z}$ or if $x\in\mathbb{R}$. If $x\in\mathbb{Z}$, there is no problem. If $x\in\mathbb{R}$ and we allow complex exponentiation, then no matter which branch cut we take, the limit does tend to $0$. – robjohn Dec 02 '19 at 16:28
  • @robjohn : I can understand your stance, this could be allowed. – MPW Dec 02 '19 at 16:42
  • @MPW Please see the update where I make it clear – gt6989b Dec 02 '19 at 17:29
  • can you please explain the meaning of "branch cut" ? @robjohn – ankit Dec 03 '19 at 05:55
  • 2
    @ankit: to define $\left(\frac1e-\frac{x}{x+1}\right)^x$ for $x\not\in\mathbb{Z}$ and $x\gt\frac1{e-1}$, we need to define how to raise a negative number to a non-integer power. The usual way is to use logarithms of negative numbers, which are complex valued: $\log(-x)=\log(x)+(2k+1)\pi i$ for some $k\in\mathbb{Z}$. The branch cut determines what $k$ is. – robjohn Dec 03 '19 at 10:24
  • Thank you. @robjohn – ankit Dec 03 '19 at 10:52
2

Assuming $f(x):\mathbb R\to \mathbb R$ the limit exists and it is equal to zero indeed according to the definition of limit (refer for example to "Rudin's Principles of Mathematical Analysis")

$$\Big(\lim_{x\rightarrow \infty} f(x) = L \Big)\iff \Big(\forall \varepsilon >0\, \exists \delta: \forall x\in D\quad x>\delta \implies \vert f(x)-L\vert <\varepsilon \Big)$$

where $D$ is the domain where $f(x)$ can be evaluated, that is

$$D = \mathbb{R} \setminus \{x:\Im(f(x))\neq0\}$$

therefore since $$\frac{1}{e} - \frac{x}{x+1} \to \frac 1e-1=\frac{1-e}{e}, \quad a=\left|\frac{1-e}{e}\right|<1 $$

then

$$\left|\frac{1}{e} - \frac{x}{x+1}\right|^{x} \to a^\infty=0$$

and therefore according to the given definition of limit, since $D$ is a not empty subset of $\mathbb R$ with $\infty$ as limit point, we have that

$$\left(\frac{1}{e} - \frac{x}{x+1}\right)^{x} \to 0$$

To a similar discussion with many good answer from many users, refer to the related

user
  • 154,566
  • 3
    I believe this is incorrect. The limit does not exist because the expression becomes undefined for arbitrarily large $x$. – MPW Dec 02 '19 at 15:35
  • 1
    @MPW I see your point, but I think that the limit exists since we can eliminate form the domain the points for which the expression in not well defined. – user Dec 02 '19 at 15:45
  • Not really. The expression is only well-defined for integral $x$. – MPW Dec 02 '19 at 15:54
  • @MPW Maybe also for some rational value. – user Dec 02 '19 at 15:57
  • I had the same thought of line as this answer, and also verified it using Wolfram|Alpha. But then Wolfram|Alpha's solution has a step without any reference, which I don't understand. Have a look at two steps above blue answer: https://imgur.com/vEbWC5d – Harshal Gajjar Dec 02 '19 at 16:04
  • See the second addendum to my answer. The answer appears to depend heavily on context and definition of existence of limit. – MPW Dec 02 '19 at 16:41
  • @MPW The limit is zero in all the cases. – user Dec 02 '19 at 18:49
  • @ParamanandSingh Sorry to disturbing you. May I have your prominent advice on the correctness or fallacy of the answer. Thanks – user Dec 02 '19 at 20:43
  • @MarkViola Sorry to disturbing you. May I have also your prominent advice on the correctness or fallacy of the answer. Thanks – user Dec 02 '19 at 20:43
  • 1
    @dfnu Why are you claiming that it is not a limit point for $D$? – user Dec 02 '19 at 21:31
  • 1
    @dfnu Indeed for that reason I'm very unhappy to see a correct answer downvoted mainly for the uncorrect first comment by MPW. Not because the answer is mine but mainl because we are giving wrong information to the asker and many other readers. – user Dec 02 '19 at 21:39
  • yep. That is a tricky-one. I am often led to confusion by the usual (and not always correct) correspondence $y=f(x)^{g(x)}=e^{g(x)\log f(x)}$. – dfnu Dec 02 '19 at 21:41
  • 1
    @dfnu Thanks to have expressed your point and discuss your doubt with me. Bye – user Dec 02 '19 at 21:47
  • 1
    @user: "where $f(x)$ can be evaluated" To evaluate $f(x)$, we must travel outside the real line, into the complex plane. If this is okay, then, yes, the limit exists and is $0$. However, if we cannot leave the real line, then $f(x)$ cannot be evaluated and so the limit doesn't exist. – robjohn Dec 02 '19 at 22:16
  • @robjohn Thanks for you kind exaplanation. Yet I can't see the point I'm loosing, maybe trivial. If we exclude from the domain the points fro which the expression is not well defined, why cannot be the limit evaluated on the real line? Is not a cse similar to the one I've linked? – user Dec 02 '19 at 22:21
  • 2
    @user: you need to define where $x$ lives. Using the variable $x$ rather than $n$ hints that the domain is $\mathbb{R}$ rather than $\mathbb{Z}$. I don't think that the limit in your cited example exists since there are holes in the domain where the function is not defined, although the function matches a function without holes that does have the limit of $1$. – robjohn Dec 02 '19 at 22:46
  • @robjohn Ok if we assume that the limit doesn't exist in th elinked example I also can understand why we are claiming that also this one doesn't exists but it seems to me that according to the given definition we can claim that the limit exists. This is of course also a matter of definition but I don't think that Rudin is a bad source on this topic. – user Dec 02 '19 at 22:50