Show boundedness of $\lim\limits_{n \to \infty}(1+\frac{x}{n})^n$

Question

Show that there is an upper boundary to $\lim\limits_{n \to \infty}\left(1+\frac{x}{n}\right)^n$, where $x\geq -n$ and $x \neq0$.

We are allowed to use $\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e^1$ and that $e^1<3$.

My approach:

$\lim\limits_{n \to \infty}\left(1+\frac{x}{n}\right)^n= \lim\limits_{n \to \infty}\left(\left(1+\frac{1}{\frac{n}{x}}\right)^{\frac{n}{x}}\right)^x$. Now I am not sure about my next steps:

As $n \to \infty$ it follows that $\frac{n}{x} \to \infty$. Hence, $\lim\limits_{\frac{n}{x} \to \infty}\left(\left(1+\frac{1}{\frac{n}{x}}\right)^{\frac{n}{x}}\right)^x=\left(\lim\limits_{\frac{n}{x} \to \infty}\left(1+\frac{1}{\frac{n}{x}}\right)^{\frac{n}{x}}\right)^x$, due to continuity of power functions we can put the $\lim$ into the brackets. Finally, we get $(e^1)^x$. So $3^x$ might serve as an upper boundary.

I am not sure if my algebraic manipulations are legit.

Any help or comments are appreciated.

Answer 1

Your solution is straightforward and elegant; it only requires a small modification. As mentioned by Tuvasbien in the comments, your proof does make an assumption that you're not given, i.e. convergence (as a continuous limit):$$ \lim_{t\rightarrow\infty} \left(1+\frac1t\right)^t = e $$ Of course, you only need to show an upper bound. To show a bound, suppose we call $m=\lfloor{t}\rfloor$: $$ \left(1+\frac1t\right)^t \le \left(1+ \frac1m\right)^t \le \left(1+ \frac1m\right)^{m+1} = \left(1+ \frac1m\right)^m \left(1+\frac1m\right) $$ and since $m$ is integer, you can use the fact that $\left(1+ \frac1m\right)^m\rightarrow e$, and you get the bound you mentioned, but more rigorously. This same approach can also give a lower bound, so you do in fact get convergence $\lim_{t\rightarrow\infty} \left(1+\frac1t\right)^t = e$, though this isn't necessary to solve the problem.

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As an aside, a perhaps more natural example of failure of extending a discrete limite to a continuous one: Take $f(x) = x\sin(\pi x)$. Then as a discrete limit ($n\in\mathbb{N}$):$$ \lim_{n\rightarrow\infty} f(n) = 0 $$ but on the other hand, as a function of a real value, $f(x)$ is not even bounded.

Answer 2

The fact that $(1+\frac1n)^n$ converges to $e<3$ implies that every term $(1+\frac1n)^n$ with $n$ large enough is bounded by $3$. To use this fact, you need to introduce things that look like $(1+\frac1n)^n$, where the exponent $n$ is an integer. To do this, you can use inequalities involving the floor function: $$ \lfloor x\rfloor \le x \le \lfloor x \rfloor + 1$$ First consider the case $x>0$. You can write $n=\frac nx\cdot x\le(\lfloor \frac nx \rfloor + 1)x=\lfloor \frac nx \rfloor x + x$ and therefore $$\left(1+\frac xn\right)^n\le \left(1+\frac 1{\lfloor n/x\rfloor}\right)^n\le\left(1+\frac1{\lfloor n/x\rfloor}\right)^{\lfloor n/x\rfloor x}\cdot\left(1+\frac1{\lfloor n/x\rfloor}\right)^x.$$ Since every occurrence of $\lfloor n/x\rfloor$ is an integer, for all large $n$ you can bound the first factor on the RHS by $3^x$. (What do you do with the second factor?)

For the case $x<0$, the above argument fails, but it's not hard to find a simple upper bound.

Answer 3

The condition $x\geq - n$ is not needed here. For any fixed value of $x$ the sequence $a_n=(1+(x/n))^n$ is bounded.

One can prove it directly using a little bit of algebra. If $x\leq 0$ then after a certain point $0\leq (1+(x/n)\leq 1$ and hence $0\leq a_n\leq 1$.

For $x>0$ we can see via binomial theorem that $$0<a_n\leq \sum_{k=0}^{n}\frac{x^k}{k!}$$ and the RHS converges to a fixed value as $n\to\infty $ so that the sequence is bounded.

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Your approach is fine but needs a little more effort as mentioned in other answers.

You can try to prove that the sequence in question converges and hence bounded. Using the limit $(1+(1/n))^n\to e$ we can prove via algebraic manipulation that the sequence converges if $x$ is an integer. If $x$ is not an integer then it lies between two integers $m$ and $m+1$ and the sequence in question thus lies between two two bounded sequences $(1+(m/n))^n$ and $(1+(m+1)/n)^n$ and therefore itself bounded.