for $z\in\mathbb C-\{0\},~\dfrac{1}{1+nz}\to0.$ How to show that for $z\in\mathbb C-\{0\},~\dfrac{1}{1+nz}\to0.$
I've tried triangle inequality couldn't arrive at any conclusion.
Please help me.
 A: Presumably you mean as $n \to \infty$?
The triangle inequality is your friend. In particular, $|1+nz| \ge n|z|-1$.
If you want more:

 Let $\epsilon>0$ and choose $N > \frac{1}{|z|} (1+\frac{1}{\epsilon})$. Then if $n \ge N$, we have$|\frac{1}{1+nz}| = \frac{1}{|1+nz|} \le \frac{1}{n|z|-1}< \frac{1}{\frac{1}{\epsilon}} = \epsilon$.

A: You can calculate
\begin{eqnarray}
0 & = & 0 \cdot \frac{1}{z} = 0 \cdot \frac{1}{0+z} = \lim_{n \rightarrow \infty} \frac{1}{n} \frac{\lim_{n \rightarrow \infty} 1}{\lim_{n \rightarrow \infty} \frac{1}{n} + \lim_{n \rightarrow\infty} z} \\
& = & \lim_{n \rightarrow \infty} \frac{1}{n} \frac{\lim_{n \rightarrow \infty} 1}{\lim_{n \rightarrow \infty} (\frac{1}{n} + z)} = \lim_{n \rightarrow \infty} \frac{1}{n} \lim_{n \rightarrow \infty} \frac{1}{\frac{1}{n} + z} \\
& = & \lim_{n \rightarrow \infty} \frac{1}{n} \frac{1}{\frac{1}{n} + z} = \lim_{n \rightarrow \infty} \frac{1}{1+nz} \ .
\end{eqnarray}
Not that in the equations with limits on both sides the existence and equality of every limit on the right hand side is always established by the existence of limits on the left hand side and limit computation rules. This is the technique that is applied in the calculation of limits of rational functions at infinity. Note that a rational function might have singularities without disturbing the existence of a limit at infinity.
More strictly speaking we have to remove the finite set of zeros of the denominator from the definition of domain of the sequence before taking limits. Then the expression $\frac{1}{1+nz}$ defines a sequence $\big\{\frac{1}{1+nz}\big\}_{n \in D} \ $, where $D \subset \mathbb{N}$, that is, a mapping $a: D \rightarrow \mathbb{C}, \ a(n) = \frac{1}{1+nz}$. Then there is a number $a^*$ such that for every $\epsilon > 0$ there is $N \in \mathbb{N}$ s.t. $n \in D$, $n > N$ implies
\begin{equation}
|a(n) - a^*| < \epsilon \ .
\end{equation}
The number $a^*$ is called the limit of the sequence $a$. In the same way we can find that $\lim_{n \rightarrow \infty} \frac{i^k}{\cos(\frac{n\pi}{2})} = 1$, that is found by removing zeros of the denominator at points $\{2i+1\}_{i=0}^\infty$ from the definition of domain of the sequence. In a similar way we obtain $\lim_{t \rightarrow n} a(t) = a(n)$ for every $n \in D$ by restricting the evaluation of the inequality $|a(n)-a^*| < \epsilon$ to the definition of domain $D$ that $a$ is a continuous mapping.
