For $\alpha \in (0,\frac{3}{2})$, define $x_n=(n+1)^{\alpha}-n^{\alpha}$. Then $\lim_{n \to \infty}x_n$ is? 
For $\alpha \in (0,\frac{3}{2})$, define $x_n=(n+1)^{\alpha}-n^{\alpha}$. Then $\lim_{n \to \infty}x_n$ is?
A) 1 for all $\alpha$
B) 1 or 0 depending on the value of $\alpha$
C) 1 or $\infty$ depending on the value of $\alpha$
D)  1 , 0 or $\infty$ depending on the value of $\alpha$

Now here I am able to get 1 and $\infty$ depending on the value of $\alpha$ but how does 0 counts in? Since it's possible to get 0 only when value of $\alpha$ is 0 but here here 0 is not included in interval.
What am I doing wrong?
 A: We can evaluate the limit a number of ways.  Herein, we present three approaches.

METHODOLOGY $1$:  Apply Standard Inequalities and the Squeeze Theorem

PRIMER:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function and the logarithm function satisfy the inequalities
$$1+x\le e^x\le \frac{1}{1-x} \tag 1$$
for $x<1$ and 
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 2$$
for $x>0$.

First we write 
$$\begin{align}
(n+1)^\alpha-n^\alpha&=e^{\alpha\log(n)}\left(e^{\alpha \log\left(1+\frac1n\right)}-1\right)\tag3
\end{align}$$
Next, we use $(1)$ and $(2)$ in $(3)$ to obtain
$$\frac{\alpha n^{\alpha-1}}{1+1/n}\le (n+1)^\alpha-n^\alpha\le \frac{\alpha n^{\alpha-1}}{1-\alpha/n}\tag 4$$
whereupon applying the squeeze theorem to $(4)$ reveals
$$\begin{align}
\lim_{n\to \infty }\left((n+1)^\alpha-n^\alpha\right)&= \begin{cases}\infty&,1<\alpha<3/2\\\\
1&,1=\alpha\\\\
0&\alpha<1
\end{cases}
\end{align}$$

METHODOLOGY $2$:  Asymptotic Analysis
Here, we expand $(n+1)^\alpha$ using the general binomial theorem to find
$$\begin{align}
(n+1)^\alpha-n^\alpha&=n^\alpha\left(\left(1+\frac1n\right)^\alpha-1\right)\\\\
&=n^\alpha\left(\left(1+\frac\alpha n+O\left(\frac1{n^2}\right)\right)-1\right)\\\\
&=\alpha n^{\alpha -1}\\\\
&\to \begin{cases}\infty&,1<\alpha<3/2\\\\
1&,1=\alpha\\\\
0&\alpha<1
\end{cases}
\end{align}$$
as expected!

METHODOLOGY $3$:  Apply the Mean Value Theorem
Finally, applying the mean value theorem to the function $f(x)=x^\alpha$ yields
$$(x+1)^\alpha-x^\alpha=\alpha \xi^{\alpha-1}$$
for $x<\xi<x+1$.  Letting $x\to \infty$, we find again that 
$$\begin{align}
\lim_{x\to \infty}\left((x+1)^\alpha-x^\alpha\right)&= \begin{cases}\infty&,1<\alpha<3/2\\\\
1&,1=\alpha\\\\
0&\alpha<1
\end{cases}
\end{align}$$
as expected!
A: $$(n+1)^{\frac{1}{2}}-n^{\frac{1}{2}}=\frac{n+1-n}{(n+1)^{\frac{1}{2}}+n^{\frac{1}{2}}}=\frac{1}{(n+1)^{\frac{1}{2}}+n^{\frac{1}{2}}}\to0 \quad \text{as }n\to\infty$$
A: If you know the mean value theorem, then let $f(x)=x^{\alpha}.$ 
By the mean value theorem, $f(n+1)-f(n)=f'(c_n)$ for some $c_n\in(n,n+1)$. 
But $f'(x)=\alpha x^{\alpha-1}$. So we have that:
$$f(n+1)-f(n)=\alpha c_n^{\alpha-1}, c_n\in(n,n+1)$$
Consider the three cases: $\alpha<1,\alpha=1,\alpha>1$. 
I'll give you a start. 
If $\alpha>1$ then $f'(x)$ is increasing, and $f(n+1)-f(n)=f'(c_n)> f'(n)=\alpha n^{\alpha-1}$. Use this to prove $f(n+1)-f(n)\to \infty$.
