Show that $\lim_{n\to\infty}n\hbox{Log}(1+\frac{z}{n})=z$? 
Show that $\lim_{n\to\infty}n\hbox{Log}(1+\frac{z}{n})=z$ where $\text{Log}$ is the principle complex logarithm. 

I am not sure how to start this question. Could anyone please provide a hint as to how to proceed with this question?
 A: By the definition of principal branch, the function $f(z)=\hbox{Log}(z)$ is holomorphic on the slit complex plane $\mathbb{C}\backslash(-\infty,0]$. Note that for any $z\in\mathbb{C}$, $1+\frac{z}{n}\in\mathbb{C}\backslash(-\infty,0]$ for large enough $n$. 
Now, by the definition of complex derivatives, 
$$
\lim_{n\to\infty}n\hbox{Log}(1+z/n)=z\lim_{n\to\infty}\frac{\hbox{Log}(1+z/n)-\hbox{Log}(1)}{z/n}
=z\cdot\frac{d}{dw}\hbox{Log}(w)\bigg|_{w=1}=z\cdot\frac{1}{w}\bigg|_{w=1}=z
$$
A: You can use a Taylor expansion:
$$ n \, \log(1 + \frac z n) = n \, \left(\underbrace{\log(1)}_{=1} + \underbrace{\log'(1)}_{=1}\,\frac z n + O\left(\left(\frac z n\right)^2\right)\right) = z + n \, O\left(\left( \frac z n \right)^2 \right) \to z $$
A: (1+(1/x))^x approaches e for x tending to infinity.
Thus ln((1+(1/x))^x) approaches 1 for x tending to infinity.
Replace x by n/z.
Thus Ans=nln(1+(z/n))=zln((1+(z/n))^(n/z)) 
For n tending to infinity, the logarithm term approaches 1.
Thus Ans=z
A: Definition:
$$
\log a = \int_1^a\frac{dx}{x}
$$
So:
$$
n\log\left(1+\frac{z}{n}\right) = n\int_1^{1+z/n} \frac{dx}{x}
\\ =
z\;\cdot\;\frac{1}{(1+z/n)-(1)}\int_1^{1+z/n}\frac{dx}{x}
$$
By continuity of $1/x$ at the point $1$,
$$
\lim_{n}n\log\left(1+\frac{z}{n}\right) =
z\cdot \frac{1}{1} = z
$$
