Math Help - Greatest integer function limit?

1. Greatest integer function limit?

Evaluate: $\lim_{x\rightarrow 0}\left(\frac{(1 + [x])^{\frac{1}{[x]}}}{e}\right)^{\frac{1}{[x]}}$
if it exists where $[x]$ denotes the greatest integer function.

2. Nonsensical sassafras?

I don't understand the point of the question. [x] is the step function, which is discontinuous at x=0. So the left hand limit does not exist, and the right hand limit would simply be the value of the function at x=1 (namely, $\frac2e$). On the other hand, $\lim_{x\rightarrow0}\left(\frac{(1+x)^{\frac1x}}{e }\right)^{\frac1x}\approx.60659132$

3. The answer in the text is 0.

4. Media Man is mistaking, the greatest integer function does exist at x=0 however it is discontinuis since it equals -1 when approached from the right and 0 when approached from the left. I think in your question you would just plug in 0 for [x]

the greatest integer function is synonmous with the floor function in computer programing
http://en.wikipedia.org/wiki/Floor_function

5. Clarity?

I think we need some clarification here. I am taking the "greatest integer function" to be the next integer up from a non-integer, or more formally: $[x]=[n+\epsilon]=n+1$ for all $n\in\mathbb{Z},\epsilon\in(0,1]$, in which case $[x]=0$ from the left and $[x]=1$ from the right, and our function is only defined on the right, in the interval $(0,1]$.

If instead we are taking $[x]=[n+\epsilon]=n$ for all $n\in\mathbb{Z},\epsilon\in[0,1)$, then yes, $[x]=-1$ from the left and $[x]=0$ from the right, so our function is completely undefined in the interval $[-1,1)$.

Either way, there is no interpretation here. (By the way, LaTex allows the use of $\lfloor x \rfloor$ and $\lceil x \rceil$ to distinguish between the two step functions.)

6. Consider when $0 \le x < 1$ then we have

$1+ [x] = 1$

so

$(1+[x])^{\frac{1}{[x]}} = 1 .$

Hence the limit becomes

$\lim_{x \to 0} e^{- \frac{1}{x}} = 0$ .

7. Originally Posted by Media_Man
I think we need some clarification here. I am taking the "greatest integer function" to be the next integer up from a non-integer, or more formally: $[x]=[n+\epsilon]=n+1$ for all $n\in\mathbb{Z},\epsilon\in(0,1]$, in which case $[x]=0$ from the right and $[x]=1$ from the left, and our function is only defined on the right, in the interval $(0,1]$.

If instead we are taking $[x]=[n+\epsilon]=n$ for all $n\in\mathbb{Z},\epsilon\in[0,1)$, then yes, $[x]=-1$ from the right and $[x]=0$ from the left, so our function is completely undefined in the interval $[-1,1)$.

Either way, there is no interpretation here. (By the way, LaTex allows the use of $\lfloor x \rfloor$ and $\lceil x \rceil$ to distinguish between the two step functions.)
The greatest integer function, $[x]$ , is the greatest integer less than or equal to $x$.

8. I must say I find all this very confusing, and it shows why names are sometimes to be preferred to symbols.
Either we are talking about the floor function or the ceiling function.
If [x] is the ceiling function [x] is 1 everywhere in (0,1) and (1+[x])^[x] = 2^1 = 2 everywhere and so the answer is 2/e.
If we mean the floor function then we are being asked to evaluate 1^infinity, which I suppose is 1, then we want (1/e)^infinity which is 0.
I don't understand how "[x]=-1 from the right" can make any kind of sense. It would make more sense the other way round. I suppose Media_Man means that [x]=-1 if you head right from the left of 0 and [x] is again the floor function.
On the left (1+[x]) is then 0 and 0^-1 is infinity. infinity/e is still infinity and infinity^-1 is 0.
So bizarrely you could say the function is continuous on both sides at 0 if the floor function is intended!

9. Originally Posted by the_doc
Consider when $0 \le x < 1$ then we have

$1+ [x] = 1$

so

$(1+[x])^{\frac{1}{[x]}} = 1 .$

Hence the limit becomes

$\lim_{x \to 0} e^{- \frac{1}{x}} = 0$ .
There again, this last line changes the problem. The limit should be $\lim_{x \to 0} e^{- \frac{1}{[x]}}$ not $\lim_{x \to 0} e^{- \frac{1}{x}}$, the difference being that $[x]$ is constant on the interval $[0,1)$, so $\lim_{x \to 0} e^{- \frac{1}{[x]}}$ = $\lim_{x \to 0} e^{- \frac{1}{0}}$ which has no meaning.

As for the first line, $[x]=0$ when $0 \le x < 1$, $(1+[x])^{\frac1{[x]}}=(1+0)^{\frac10}$ . How does this equal 1?

$[x]$ is a constant function on small intervals, so $\lim_{x \to 0^+} f([x])=f(0)$ and $\lim_{x \to 0^-} f([x])=f(-1)$, so the limit only exists if $f(0)$ and $f(-1)$ both exist and equal one another.

10. Originally Posted by alunw
On the left (1+[x]) is then 0 and 0^-1 is infinity. infinity/e is still infinity and infinity^-1 is 0.
So bizarrely you could say the function is continuous on both sides at 0 if the floor function is intended!
The red above is absolutely wrong: $0^{-1}$ is undefined.
In all of the replies, Media_Man has been correct.
For all $x \in [ - 1,0)$ the value of the function at $x$ is undefined because $0^{-1}$ is undefined.
Therefore the limit from the left at $x=0$ is undefined.

11. I don't understand how "[x]=-1 from the right" can make any kind of sense.
Alunw: This was a typo I just corrected. According to the the_doc, [x] is referring to what I hold in my mind to be $\lfloor x \rfloor$, the "floor function".

Consider this problem as the composition of functions: $\lim_{x \to 0} (f\circ g)(x)$, where $f(x)=\left(\frac{(1+x)^{\frac1x}}{e}\right)^{\frac 1x}$ and $g(x)=\lfloor x \rfloor$. We then see that $\lim_{x\to 0^+} (f\circ g)(x)=\lim_{g(x)=0} f(g(x))=f(0)$ and $\lim_{x\to 0^-} (f\circ g)(x)=\lim_{g(x)=-1} f(g(x))=f(-1)$

If we want to talk philosophy instead of math, we could easily say $f(0)=\left(\frac{(1+0)^{\frac10}}{e}\right)^{\frac 10}=\left(\frac{1^{\infty}}{e}\right)^{\infty}=0$ and $f(-1)=\left(\frac{(1-1)^{\frac1{-1}}}{e}\right)^{\frac1{-1}}=\left(\frac{0^{-1}}{e}\right)^{-1}=0$ . But rest assured, this is not mathematics. $f(\infty)$ only has rigorous meaning as $\lim_{x\to\infty} f(x)$, which is not the case in this problem.

12. 0^-1 undefined?

Originally Posted by Plato
The red above is absolutely wrong: $0^{-1}$ is undefined.
I'll happily concede that one should not use infinity as cavalierly as I did in my last post. However, I don't think I was "absolutely wrong". I'd say 0^-1 =1/0 and there are contexts (e.g. Complex analysis done on the Riemann sphere) where one can assign a sensible value to 0^-1 as there is only one infinity.
It makes much less sense here because we must be talking about real functions, and 0^-1 is undefined since it tends to -infinity on the left and +infinity on the right.

13. I'm glad you posted your correction as I was starting to think there was some other kind of greatest integer function or that I didn't know what right and left meant any more.

As for the limit once we split the functions as you suggest to get rid of the floor function then your f(x) is not defined at 0. But I think it has a sensible limit as x->0 from the right which you already found numerically and which I think is (1/e)^0.5, though I just did that by trial and error on a calculator. So if you define f at 0 by defining it to be (1/e)^0.5 then the limit of the composed functions would have to be that!

14. Originally Posted by Media_Man
There again, this last line changes the problem. The limit should be $\lim_{x \to 0} e^{- \frac{1}{[x]}}$ not $\lim_{x \to 0} e^{- \frac{1}{x}}$, the difference being that $[x]$ is constant on the interval $[0,1)$, so $\lim_{x \to 0} e^{- \frac{1}{[x]}}$ = $\lim_{x \to 0} e^{- \frac{1}{0}}$ which has no meaning.

As for the first line, $[x]=0$ when $0 \le x < 1$, $(1+[x])^{\frac1{[x]}}=(1+0)^{\frac10}$ . How does this equal 1?

$[x]$ is a constant function on small intervals, so $\lim_{x \to 0^+} f([x])=f(0)$ and $\lim_{x \to 0^-} f([x])=f(-1)$, so the limit only exists if $f(0)$ and $f(-1)$ both exist and equal one another.
By this I meant that the limits are the same - which they are. Why should this have no meaning? I see nothing in the definition of a limit that says that a constant function cannot have a limit.

$\lim_{x \to 0} 1 = 1$,

does not seem to contradict the definition of a limit in Spivak:

The function $f$ approaches the limit $l$ near $a$ means: for every $\epsilon > 0$ there is some $\delta > 0$ such that, for all $x$, if $0 < |x-a| < \delta$ , then $|f(x) -l| < \epsilon$ .
1 to the power of anything is 1.

Moreover, I believe it is implied that this is a one sided limit from the right as
$[x]$ is discontinuous as $x$ approaches $0$ from the left!

15. 1 to the power of anything is 1.
But if something which has a limit 1 is taken to a given power, then the limit is not necessarily 1.
I don't know if you're aware of that, as I haven't read all the previous posts lol

Example ; $\left(1+\frac 1n\right)^n$

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