Evaluate $\lim \limits_{n \to \infty\ } n((n+1)^{\frac{1}{100}}-n^\frac{1}{100})$ $\lim \limits_{n \to \infty\ } n((n+1)^{\frac{1}{100}}-n^\frac{1}{100})$
$n((n+1)^{\frac{1}{100}}-n^\frac{1}{100})=n^\frac{101}{100}((1+\frac{1}{n})^{\frac{1}{100}}-1)=n^\frac{101}{100}(e^{\frac{ln(1+1/n)}{100}-}-1)$...
hmm, Is this a good approach?
 A: hint
$$(n+1)^a-n^a=n^a\Bigl((1+\frac{1}{n})^a-1\Bigr)$$
observe that
$$\lim_{n\to +\infty}\frac{e^{a\ln(1+\frac 1n)}-1}{a\ln(1+\frac 1n)}=1$$
and
$$\lim_{n\to+\infty}n\ln(1+\frac 1n)=1$$
You will find $+\infty$.
A: Hint:
Rationalize the numerator
The general term of the denominator will be $$(n+1)^{(99-r)/100}n^{r/100},0\le r\le99$$
Alternatively,
Set $1/n$ to find 
$$\lim_{n \to \infty } n((n+1)^m-n^m=\lim_{h\to0^+}\dfrac{(1+h)^m-1}{h^{1+m}}=\lim_{h\to0^+}\dfrac{mh+\binom m2h^{m-1}+\cdots+h^m}{h^{1+m}}=+\infty$$
A: Use binomial expansion
$$(n+1)^{\frac{1}{100}}=n^\frac{1}{100}(1+1/n)^{\frac{1}{100}}=n^\frac{1}{100}\left(1+\frac1{100n}+O\left(\frac1{n^2}\right)\right)=$$$$=n^\frac{1}{100}+\frac1{100n^{99/100}}+O\left(\frac1{n^{199/100}}\right)$$
and therefore
$$ n((n+1)^{\frac{1}{100}}-n^\frac{1}{100})=\frac{n^{\frac1{100}}}{100}+O\left(\frac1{n^{99/100}}\right)\to\infty$$
A: Depending on what exactly you're able to use, this certainly could be a good approach. If you know such facts as
$$\ln\left(1 + \frac 1 n\right) \ge \frac 1 2 \cdot\frac 1 n$$
for all sufficiently large $n$ (which is an easy consequence of the fact that $\ln(1+x)$ has derivative $1$ at $x = 0$), then you can quickly finish the problem:
$$e^{\frac{\ln(1+1/n)}{100}} - 1 \ge e^{1/(200n)} - 1 \ge 1 + \frac{1}{200 n} - 1 = \frac{1}{200n}.$$
where the inequality is a consequence of the fact that $e^{t} \ge 1 + t$. Now combine this with your power of $n$ to see that the limit is infinite.
That being said, I certainly think that binomial expansion is a more natural approach to the problem. But the advantage of your method is that it really only requires two facts: that the derivative of $\ln(1+x)$ is $1$ when $x$ is zero, and that the derivative of $\exp$ is $1$ when $x$ is zero (plus, of course, the mean value theorem to actually apply this information).
