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.
 A: Thanks this answer's here : Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$
Consider
$$
f(x)=[( 1+x^{p+1})^{\frac{1}{p+1}}-(1+x^p)^{\frac1p}]
$$
Then
$$
f'(x)=x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]
$$

If $p>1$ therefore $f'(0)=0$. Therefore
  $$
0=f'(0)=\lim_{x\to0^+}x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]=
\lim_{t\to\infty}\bigl([( 1+t^{p+1})^{\frac{1}{p+1}}-(1+t^p)^{\frac1p}]\bigr)
$$
  with the substitution $t=1/x$.
If $0<p<1$ therefore $f'(0)=-\infty$. Therefore
  $$
-\infty =f'(0^+)=\lim_{x\to0^+}x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]=
\lim_{t\to\infty}\bigl([( 1+t^{p+1})^{\frac{1}{p+1}}-(1+t^p)^{\frac1p}]\bigr)
$$
  with the substitution $t=1/x$.
If $p=1$ therefore $f'(0)=-1 $. Therefore
  $$
-1 =f'(0)=\lim_{x\to0^+}x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]=
\lim_{t\to\infty}\bigl([( 1+t^{p+1})^{\frac{1}{p+1}}-(1+t^p)^{\frac1p}]\bigr)
$$
  with the substitution $t=1/x$.

A: 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 $-\infty$.

A: For any $x>0$ and $\alpha>0$ we clearly have $(1+x^\alpha)^{\frac{1}{\alpha}}\geq x$. If $\alpha=p\in\mathbb{N}^+$,
$$ (1+x^p)^{\frac{1}{p}}=\text{GM}\left[\underbrace{x,\ldots,x}_{p-1\text{ times}},x+x^{1-p}\right]\leq \text{AM}\left[\underbrace{x,\ldots,x}_{p-1\text{ times}},x+x^{1-p}\right]=x+\frac{1}{px^{p-1}}$$
and $\left(x+\frac{1}{px^{p-1}}\right)^p\geq x^p+1$, for any $p>1$, can also be seen as a consequence of Bernoulli's inequality. It follows that the difference between $ (1+x^p)^{\frac{1}{p}}$ and $ (1+x^{p+1})^{\frac{1}{p+1}}$ goes to zero as $x\to +\infty$.
It just remains to study the case $p\in(0,1]$.
A: The question says $p>0$. Let $x=\frac{1}{t}$. $$\lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}]=$$
$$=\lim_{t\to 0^+}\frac{(t^{p+1}+1)^{\frac{1}{p+1}}-1}{t}-\lim_{t\to 0^+}\frac{(t^p+1)^{\frac{1}{p}}-1}{t}=$$
$$=((t^{p+1}+1)^{\frac{1}{p+1}})'|_{t=0}-((t^p+1)^{\frac{1}{p}})'|_{t=0^+}=$$
$$=\left(\frac{1}{p+1}(t^{p+1}+1)^{-\frac{p}{p+1}}(p+1)t^p\right)|_{t=0}-$$
$$-((t^p+1)^{\frac{1}{p}})'|_{t=0^+}=0-((t^p+1)^{\frac{1}{p}})'|_{t=0^+}=$$
$$=-((t^p+1)^{\frac{1}{p}})'|_{t=0^+}$$
If $p>1$, then
$$=-\left(\frac{1}{p}(t^p+1)^{\frac{1-p}{p}}pt^{p-1}\right)|_{t=0^+}=-0=0$$
If $0<p<1$, then
$$=-\left(\frac{1}{p}(t^p+1)^{\frac{1-p}{p}}pt^{p-1}\right)|_{t=0^+}=-\infty$$
If $p=1$, then
$$=-(t+1)'|_{t=0^+}=-1$$
A: Let try to use a general method to solve this kind of limits by
looking at the first order Taylor expansion of your expression:
\begin{aligned}\left( 1+x^{p+1}\right)^{\frac{1}{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}} \\
&= x\left[ \left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- \left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}}\right] \\
&= x\left[1+\frac{1}{x^{p+1}{p+1}}+o(x^{-p-1}) - 1 -\frac{1}{x^{p}{p}}+o(x^{-p})\right]\\
&= x\left[\frac{1}{x^{p+1}{p+1}} -\frac{1}{x^{p}{p}}+o(x^{-p})\right]\\
&=\frac{1}{x^{p}{p+1}} -\frac{1}{x^{p-1}{p}}+o(x^{-p+1})
\end{aligned}
So that the whole expression tends to 0 when $x\rightarrow \infty$.
