How to compute the following limit$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$ I am trying to find
$$  \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$$
where $p>0$. I have tried to factor out as
$$(1+x^{p+1})^{\frac1{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} =x\left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- x\left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}},$$
 but still was not able to make progress. Any other approach to this is welcome.

We can use the expansion
$$(1+y)^{\alpha} = \sum_{n=0}^{\infty} \binom{\alpha}{n} \cdot y^n.$$
Taking $y = \frac{1}{x^p}, \ \alpha = \frac{1}{p}$, we get 
$$\left( 1 + \frac{1}{x^p} \right)^{\frac{1}{p}} = 1 + \frac{1}{p x^p} + o \left( \frac{1}{x^{p}} \right).$$
Similarly
$$\left(1+\frac{1}{x^{p+1}} \right)^{\frac{1}{p+1}} = 1+\frac{1}{(p+1)x^{p+1}} + \mathcal{o} \left( \frac{1}{x^{p+1}} \right) = 1 + o \left( \frac{1}{x^p} \right).$$
Hence
$$\begin{align*}
\left( 1+x^{p+1} \right)^{\frac{1}{p+1}} - \left(1+x^p\right)^{\frac{1}{p}} & = x \left[ \left( 1 + \frac{1}{x^p} \cdot o(1) \right) - \left( 1 + \frac{1}{x^p} \left( \frac{1}{p} + o(1) \right) \right) \right] \\[1ex]
& = -\frac{x}{x^p} \left( \frac{1}{p} + o(1) \right).
\end{align*}$$


*

*If $p > 1$, the limit is $0$.

*If $p = 1$, the limit is $-1$.

*If $p < 1$, the limit is $-\infty$.

