How to find $\lim_{n \to \infty}|\left( 1+\frac{z}{n}\right)^{n}|$? This is a step on the way to proving 
$$
\lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n} = e^{z}.$$

I'm looking for an answer without 
1) a summation or 
2) a logarithmic function. 

So far, I have that 
$$
\lim_{n \to \infty} |z_{n}| = \lim_{n \to \infty} \vert\left(1+\frac{z}{n}\right)^{n}\vert = \lim_{n \to \infty} \left(\sqrt{\left(1+\frac{x}{n}\right)^{2} +\left( \frac{y}{n}\right)^{2} }\right)^{n} = \lim_{n \to \infty}\left(\left( 1+\frac{x}{n}\right)^{2} + \left(\frac{y}{n} \right)^{2} \right)^{n/2} = \lim_{n \to \infty} \left( 1 + \frac{2x}{n} + \left(\frac{x}{n}\right)^{2} + \left( \frac{y}{n}\right)^{2}\right)^{n/2}.
$$
Now, I was given the following hint, and I would like to use it, but I don't know how:


  Hint: Argue that you can discard terms with $1/n^{2}$; for example, by showing that the limit $\displaystyle \lim_{n \to \infty} \frac{\left(\left( 1 + \frac{x}{n}\right)^{2} + \left(\frac{y}{n}\right)^{2}\right)^{n/2}}{\left(1 + \frac{2x}{n}\right)^{n/2}} = 1$.


How would I show this ($\epsilon$-definition?), and how would it help me here with what I've already done?
I think the best approach is the one which I found on MSE (will add link when I find it). It is based on the following general theorem.


  Theorem: If $a_{n}$ is a sequence of real or complex terms such that $n(a_{n} - 1) \to 0$ as $n \to \infty$ then $a_{n}^{n}  \to 1$ as $n \to \infty$.


The proof is done by writing $a_{n} = 1 + a_{n} - 1 = 1 + b_{n}$ (say) so that $nb_{n} \to 0$ as $n \to \infty$. Clearly we have
\begin{align}
a_{n}^{n} &amp;= (1 + b_{n})^{n}\notag\\
&amp;= 1 + nb_{n} + \dfrac{1 - \dfrac{1}{n}}{2!}(nb_{n})^{2} + \cdots\notag
\end{align}
and hence
\begin{align}
|a_{n}^{n} - 1| &amp;\leq |nb_{n}| + \frac{1}{2!}|nb_{n}|^{2} + \cdots + \frac{1}{n!}|nb_{n}|^{n}\notag\\
&amp;\leq|nb_{n}| + \frac{1}{2}|nb_{n}| + \frac{1}{2^{2}}|nb_{n}|^{2} + \cdots\notag\\
&amp;= \dfrac{|nb_{n}|}{1 - \dfrac{|nb_{n}|}{2}}\notag\\
&amp;\to 0\text{ as }n \to \infty\notag
\end{align}
Hence $a_{n}^{n} \to 1$ as $n \to \infty$.

It is now easy to show that if $z = x + iy$ then $$\left(1 + \frac{z}{n}\right)^{n} \to e^{x}(\cos y + i\sin y)$$ for all real $x, y$ where symbol $e^{x}$ is defined by the limit $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{1}$$ I assume that you know the proof that the limit in $(1)$ exists for all real $x$ and this limit is non-zero.

I show that if $$f(z) = \lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n}\tag{2}$$ then $$f(z_{1} + z_{2}) = f(z_{1})f(z_{2})\tag{3}$$ for all $z_{1}, z_{2} \in \mathbb{C}$. Consider the sequence $$a_{n} = \dfrac{1 + \dfrac{z_{1} + z_{2}}{n}}{\left(1 + \dfrac{z_{1}}{n}\right)\left(1 + \dfrac{z_{2}}{n}\right)}\tag{4}$$ and then $$n(a_{n} - 1) = -\left(\dfrac{\dfrac{z_{1}z_{2}}{n}}{\left(1 + \dfrac{z_{1}}{n}\right)\left(1 + \dfrac{z_{2}}{n}\right)}\right)$$ so that $n(a_{n} - 1) \to 0$ as $n \to \infty$. Hence $a_{n}^{n} \to 1$ as $n \to \infty$. From definition of $a_{n}$ in equation $(4)$ it follows that $f(z_{1} + z_{2}) = f(z_{1})f(z_{2})$ provided that both the limits $f(z_{1}), f(z_{2})$ exist and are non-zero.

Next I show that the limit $f(z)$ defined by equation $(2)$ exists and is non-zero for all $z \in \mathbb{C}$. If $z$ is real then by assumption mentioned after equation $(1)$, the limit $f(z)$ exists and is non-zero. Next let $z = iy$ be totally imaginary. I will show that in this case $f(z) = \cos y + i\sin y$ and hence it is non-zero. Consider again the sequence $$c_{n} = \dfrac{1 + \dfrac{iy}{n}}{\cos\left(\dfrac{y}{n}\right) + i\sin\left(\dfrac{y}{n}\right)} = \cos\left(\frac{y}{n}\right) + \left(\frac{y}{n}\right)\sin\left(\frac{y}{n}\right) + i\left\{\left(\frac{y}{n}\right)\cos\left(\frac{y}{n}\right) -\sin\left(\frac{y}{n}\right)\right\}$$ and it is easy to see that $n(c_{n} - 1) \to 0$ as $n \to \infty$ and hence $c_{n}^{n} \to 1$ and from defining equation for $c_{n}$ we can see that denominator of $c_{n}^{n}$ is $(\cos y + i\sin y)$. It follows that $$(1 + iy/n)^{n} \to (\cos y + i\sin y)$$ Thus we have established that $f(z)$ exists and is non-zero in both the cases when $z$ is purely real and when $z$ is purely imaginary.

From equation $(3)$ it follows that if $z = x + iy$ then $f(z) = f(x)f(iy)$ and $f(z)$ is non-zero because $f(x), f(iy)$ are non-zero. We thus have $f(z) = e^{x}(\cos y + i\sin y)$.



It is easy to use the theorem mentioned in the beginning of my post to prove the result which OP is seeking in order to develop a proper theory of the limit of $(1 + z/n)^{n}$. Thus I show directly that if $z_{n} = (1 + z/n)^{n}$ then $|z_{n}|^{2} \to e^{2x}$ where $z = x + iy$ and hence $|z_{n}| \to e^{x}$. Clearly as mentioned by OP we have $$|z_{n}|^{2} = \left\{\left(1 + \frac{x}{n}\right)^{2} + \frac{y^{2}}{n^{2}}\right\}^{n}$$ Taking $$a_{n} = \dfrac{\left(1 + \dfrac{x}{n}\right)^{2} + \dfrac{y^{2}}{n^{2}}}{1 + \dfrac{2x}{n}}$$ we can see that $n(a_{n} - 1) \to 0$ and hence $a_{n}^{n} \to 1$ so that $|z_{n}|^{2} \to e^{2x}$.

OP has asked in the comments about choosing an appropriate sequence $a_{n}$ to make use of the theorem mentioned in my post. The basic idea is that the theorem is supposed to be used in situations where you want to find the limit of an expression whose exponent is $n$. Thus say you want to find limit of a sequence of type $c_{n}^{n}$. Moreover you intuitively know the answer as to what the limit will be. The idea is to find another sequence of type $d_{n}^{n}$ whose limit you already know and then set $a_{n} = c_{n}/d_{n}$.

Thus for example in OP's question there is a hint that the limit of $\left(\left(1 + \dfrac{x}{n}\right)^{2} + \dfrac{y^{2}}{n^{2}}\right)^{n/2}$ is related to that of the sequence $(1 + 2x/n)^{n/2}$. Squaring these sequences we get the exponents as $n$ and then it is a simple matter to figure out that $$c_{n} = \left(1 + \dfrac{x}{n}\right)^{2} + \dfrac{y^{2}}{n^{2}}, d_{n} = 1 + \frac{2x}{n}, a_{n} = \frac{c_{n}}{d_{n}}$$ A better case can be made for the limit $(1 + iy/n)^{n}$. We intuitively know that the limit should be $e^{iy} = \cos y + i\sin y$. Let's then try to find another expression whose $n^{\text{th}}$ power tends to $\cos y + i\sin y$. Here we are damn lucky because of De Moivre's Theorem $$\left(\cos\frac{y}{n} + i\sin\frac{y}{n}\right)^{n} = \cos y + i\sin y$$ and therefore we set $$c_{n} = 1 + \frac{iy}{n}, d_{n} = \cos\frac{y}{n} + i\sin\frac{y}{n}, a_{n} = \frac{c_{n}}{d_{n}}$$
Just show that

$$\left(1 + \frac{2x}{n} + \frac{D}{n^2}\right)^n = \sum_{k=0}^n \left(1+\frac{2x}{n}\right)^{n-k} C_{n}^k \frac{D^{k}}{n^{2k}} = (1 + \frac{2x}{n})^n + \sum_{k=1}^{n} (1 + \frac{2x}{n})^{n-k}C_n^k \frac{D^k}{n^{2k}}$$

the latter term take $n\to \infty$, there is a constant $M$ that $(1 + \frac{2x}{n})^{n-k} \le M$ for all $k$. thus we just need to show that

$$\sum_{k=1}^{n} C_n^k \frac{D^k}{n^{2k}}\to 0$$

$$\sum_{k=1}^{n} C_n^k \frac{D^k}{n^{2k}} = \sum_{k=1}^{n} \frac{n(n-1)\dots (n-k+1)}{k(k-1)\dots 1} \frac{D^k}{n^{2k}}$$

and

$$\frac{n}{k}\le \frac{n-1}{k-1}\le \dots \le \frac{n-k+1}{1}$$

we just have to show

$$\sum_{k=1}^{n}\left(n-k+1\right)^k\frac{D^k}{n^{2k}}\to 0$$

while

$$(n-k+1)\le n$$

$$\sum_{k=1}^{n}\left(n-k+1\right)^k\frac{D^k}{n^{2k}}\le \sum_{k=1}^{n} \frac{D^kn^k}{n^{2k}} = \sum_{k=1}^{n}\frac{D^k}{n^k} \le \sum_{k=1}^{\infty} \frac{D^k}{n^k} = \frac{D}{n-D}\to 0$$