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 $p&gt;0$. 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.
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 $p&gt;1$ 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 $0&lt;p&lt;1$ 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 $t=1/x$.
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}} &amp;= 
x\left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- x\left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}} \\
&amp;= 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] \\
&amp;= x\left[1+\frac{1}{x^{p+1}{p+1}}+o(x^{-p-1}) - 1 -\frac{1}{x^{p}{p}}+o(x^{-p})\right]\\
&amp;= x\left[\frac{1}{x^{p+1}{p+1}} -\frac{1}{x^{p}{p}}+o(x^{-p})\right]\\
&amp;=\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 $x\rightarrow \infty$.