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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.

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Hint: substitute $y=1+x$, now apply the binomial theorem and see the limit of individual terms. –  Dinesh Nov 10 '11 at 7:26
    
@Dinesh, The binomial theorem? Sure about this? I see the wrong proof your hint may lead to... –  Did Nov 10 '11 at 7:44
    
@Didier Piau oops! you are right, I overlooked the coefficients :| –  Dinesh Nov 10 '11 at 10:26
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3 Answers

$\rm\bf Note$: $$\frac{x^n}{(1+x)^{n-1}}=(1+x)\left(\frac{x}{1+x}\right)^n.$$ 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; $$|x|<|1+x|\implies \mathrm{Re}(x)^2<(1+\mathrm{Re}(x))^2\implies \mathrm{Re}(x)>-1/2.$$

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$$\lim_{n\rightarrow\infty}\frac{x^n}{(1+x)^{n-1}}=\lim_{n\rightarrow\infty}\frac{x\cdot x^{n-1}}{(1+x)^{n-1}}=\lim_{n\rightarrow\infty}\frac{x}{(\frac{1}{x}+1)^{n-1}}=0$$

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$\frac{x^n}{(1+x)^{n-1}} = \frac{x^n(1+x)}{(1+x)^{n}} = (\frac{x}{1+x})^{n}(1+x) = (\frac{x+1-1}{1+x})^{n}(1+x) = (1 - \frac{1}{1+x})^{n}(1+x)$

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 $(1+x)*0 = 0$

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