How would I evaluate this limit? I have no idea how to evaluate this limit. Wolfram gives $0$, and I believe this, but I would like to see how it is done. The limit is
$$\lim_{n\rightarrow\infty}\frac{x^n}{(1+x)^{n-1}}$$
assuming $x$ is positive. Thanks in advance.
 A: $\rm\bf Note$:
$$\frac{x^n}{(1+x)^{n-1}}=(1+x)\left(\frac{x}{1+x}\right)^n.$$
Now write $a=x/(1+x)$ and the limit becomes $(1+x)\lim\limits_{n\to\infty} a^n$. This is $0$ if $|a|<1$, would be $1+x$ if $a=1$ (this is impossible - try and see why), and the limit doesn't exist otherwise. You can solve the inequality $|a|<1$ in $\mathbb{C}$ for $x$ by squaring and canceling and rearranging; $$|x|<|1+x|\implies \mathrm{Re}(x)^2<(1+\mathrm{Re}(x))^2\implies \mathrm{Re}(x)>-1/2.$$
A: $$\lim_{n\rightarrow\infty}\frac{x^n}{(1+x)^{n-1}}=\lim_{n\rightarrow\infty}\frac{x\cdot x^{n-1}}{(1+x)^{n-1}}=\lim_{n\rightarrow\infty}\frac{x}{(\frac{1}{x}+1)^{n-1}}=0$$
A: $\frac{x^n}{(1+x)^{n-1}} = \frac{x^n(1+x)}{(1+x)^{n}} = (\frac{x}{1+x})^{n}(1+x) = (\frac{x+1-1}{1+x})^{n}(1+x) = (1 - \frac{1}{1+x})^{n}(1+x)$
So taking the limit:
$\lim_{n\to\infty} \frac{x^n}{(1+x)^{n-1}} = \lim_{n\to\infty} (1 - \frac{1}{1+x})^{n}(1+x) = (1+x) * \lim_{n\to\infty} (1 - \frac{1}{1+x})^{n}$
Since $x>0$ we know $1 - \frac{1}{1+x} < 1$ therefore $\lim_{n\to\infty} (1 - \frac{1}{1+x})^{n} = 0$ giving us $(1+x)*0 = 0$
