Boundedness of a function

See Hai

New member
Consider the function f(x)= (1+1/x)^x . My aim is to prove that this function is bounded and its limit exists as x approaches infinity, for all positive real x. Now, I am aware of many hand waving proofs available online that simply state that the limit of f(x) is equal to e: Such proofs merely invoke the use of L Hopital's rule and are circular in nature. Unfortunately, I have not been able to find a completely rigorous proof online(is it because the level of difficulty is too high??), so I thought it would be appropriate for me to ask my question here.
What I have managed to prove so far:
1) f(x) is monotone increasing over the domain (0,inf)
2) If we merely consider the sequence x(n)=(1+1/n)^n then I can find a upper bound of 3 and then show, by the Monotone Convergence Theorem, that the sequence converges to a unique limit, which, we define to be e.(Correct me if I am wrong, I might have left out some details here).
3) However, in the case of the continuous function f(x), I am unable to prove that its limit at infinity exists, I don't know how to find a upper bound for f(x). If I could, I am done, since I can just apply MCT. Or, alternatively, I can just show that the supremum of f(x)=e, from which we are done immediately without having to invoke MCT. So, I was just wondering: Is it possible to provide a completely rjgorigo proof of the fact that f(x) has an upper bound, or perhaps even the limit at infinity of f(x) exists? I may be a little confused here due to my relatively insufficient knowledge, so any clarification will be much appreciated.
Attached below is a scheme of thought from my teacher, who is unable to prove the last two points. I don't know if the train of thought is logical though?

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pka

Elite Member
Consider the function f(x)= (1+1/x)^x . My aim is to prove that this function is bounded and its limit exists as x approaches infinity, for all positive real?
Here is a related proof. You need to know Bernoulli's inequality.
Define $$\displaystyle e_n=\left(1+\frac{1}{n}\right)^n$$
\displaystyle \begin{align*}\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^n&=\left(1-\frac{1}{n^2}\right)^n \\&\ge 1+ n\left(\frac{-1}{n^2}\right)\text{ Bernoulli's}\\\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^n&\ge\left(1-\frac{1}{n}\right)\\\left(1+\frac{1}{n}\right)^n&\ge\left(1-\frac{1}{n}\right)^{1-n}\\&\ge\left(\frac{n}{n-1}\right)^{n-1}\\&\ge\left(1-\frac{1}{n-1}\right)^{n-1}\end{align*}
That proves that $$\displaystyle e_{n}\ge e_{n-1}$$ OR the sequence s increasing.

Now we turn to Napper's inequality. If $$\displaystyle 0<a<b$$ then $$\displaystyle \frac{1}{b}<\frac{\log(b)-\log(a)}{b-a}<\frac{1}{a}$$.
For $$\displaystyle K>2$$ take $$\displaystyle b=K~\&~a=1$$ we get $$\displaystyle 1-\tfrac{1}{K}<\log(K)<K-1$$
If $$\displaystyle K=\left(1+\tfrac{1}{n}\right)$$ then $$\displaystyle \frac{1}{n+1}<\log\left(1+\tfrac{1}{n}\right)<\frac{1}{n}$$
Then $$\displaystyle e^{\frac{n}{n+1}}<\left(1+\tfrac{1}{n}\right)^n <e$$
You should be able to fill in details and finish.

See Hai

New member
Here is a related proof. You need to know Bernoulli's inequality.
Define $$\displaystyle e_n=\left(1+\frac{1}{n}\right)^n$$
\displaystyle \begin{align*}\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^n&=\left(1-\frac{1}{n^2}\right)^n \\&\ge 1+ n\left(\frac{-1}{n^2}\right)\text{ Bernoulli's}\\\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^n&\ge\left(1-\frac{1}{n}\right)\\\left(1+\frac{1}{n}\right)^n&\ge\left(1-\frac{1}{n}\right)^{1-n}\\&\ge\left(\frac{n}{n-1}\right)^{n-1}\\&\ge\left(1-\frac{1}{n-1}\right)^{n-1}\end{align*}
That proves that $$\displaystyle e_{n}\ge e_{n-1}$$ OR the sequence s increasing.

Now we turn to Napper's inequality. If $$\displaystyle 0<a<b$$ then $$\displaystyle \frac{1}{b}<\frac{\log(b)-\log(a)}{b-a}<\frac{1}{a}$$.
For $$\displaystyle K>2$$ take $$\displaystyle b=K~\&~a=1$$ we get $$\displaystyle 1-\tfrac{1}{K}<\log(K)<K-1$$
If $$\displaystyle K=\left(1+\tfrac{1}{n}\right)$$ then $$\displaystyle \frac{1}{n+1}<\log\left(1+\tfrac{1}{n}\right)<\frac{1}{n}$$
Then $$\displaystyle e^{\frac{n}{n+1}}<\left(1+\tfrac{1}{n}\right)^n <e$$
You should be able to fill in details and finish.

Would you mind taking a look at my concluding statement?
" Since f(x) is monotone increasing over the interval (0,inf), and it has an upper bound,e, therefore by the MCT, we can conclude that it will converge to its supremum, which is the limit as x approaches infinity" ?
I believe the rest is just an exercise in algebra so I would not trouble you here.

See Hai

New member
Also, just to clarify, the substitution K=(1+1/n) can be done for all real n, right? It's not just restricted to the positive integers?