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

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.
A: 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.
A: 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. 

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. 
