Filling in details for calculation of the limit $\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right)$ I want to evaluate the following limit using asymptotics
\begin{equation}
\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) \tag{1}
\end{equation}
This is an example problem and this was the solution:
\begin{gather}
\frac{x^{3} +x}{1+x^{3}} =\left( 1+\frac{1}{x^{2}}\right)\left( 1+\frac{1}{x^{3}}\right)^{-1} =\left( 1+\frac{1}{x^{2}}\right)\left( 1-\frac{1}{x^{3}} +\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) \tag{2}\\
=1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{3}
\end{gather}
And then
\begin{gather}
\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} =\left( 1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\right)^{\frac{1}{7}} \tag{4}\\
=1+\frac{1}{7} .\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{5}
\end{gather}
\begin{equation}
\cos\frac{1}{x} =1-\frac{1}{2x^{2}} +\mathcal{O}\left(\frac{1}{x^{4}}\right) \tag{6}
\end{equation}
From above, we obtain:
\begin{equation}
\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{7}
\end{equation}
Hence the required limit is:
\begin{equation}
\lim _{x\rightarrow \infty }\left(\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\right) =\frac{9}{14} \tag{8}
\end{equation}
I tried to fill in the details for the steps involved in the above steps.
I supplied the details for $\displaystyle ( 3)$ as below:
\begin{gather}
\left( 1+\frac{1}{x^{2}}\right)\left( 1-\frac{1}{x^{3}} +\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) =1+\frac{1}{x^{2}} +\frac{1}{x^{3}}\left( -1-\frac{1}{x^{2}} +\left( x^{3} +x\right)\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) \tag{9}\\
=1+\frac{1}{x^{2}} +\frac{1}{x^{3}}\left( -1-\frac{1}{x^{2}} +\frac{\left( x^{3} +x\right)}{x^{6}}\mathcal{O}( 1)\right) =1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{10}
\end{gather}
where in (9) and (10), I have used the standard result: $\displaystyle \frac{\mathcal{O}( f( x))}{g( x)} =\mathcal{O}\left(\frac{f( x)}{g( x)}\right)$, if $\displaystyle g( x) \neq 0$.
I tried to get (5) from (4) but failed.
Then I supplied details for $\displaystyle ( 7)$ as below:
\begin{gather*}
\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) -\mathcal{O}\left(\frac{1}{x^{4}}\right) =\frac{9}{14x^{2}} +\frac{1}{x^{3}}\left(\mathcal{O}( 1) -\mathcal{O}\left(\frac{1}{x}\right)\right)\\
=\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\\
\Longrightarrow \lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\lim _{x\rightarrow \infty }\left(\frac{9}{14} +\mathcal{O}\left(\frac{1}{x}\right)\right) =\frac{9}{14}
\end{gather*}
Any help in getting (5) from (4) is much appreciated. Thanks.
\begin{equation*}
\end{equation*}
\begin{equation*}
\end{equation*}
 A: You can make life easier letting $x=\frac 1y$. This makes the expression
$$A=\frac 1{y^2} \left(\frac{\sqrt[7]{1+y^2} }{\sqrt[7]{1+y^3} } - \cos(y)\right)$$  Now, using the binomial theorem or Taylor series for the surds and Taylor series for the cosine, you have
$$A=\frac 1{y^2} \left(\frac{1+\frac{y^2}{7}-\frac{3 y^4}{49}+O\left(y^6\right)}{1+\frac{y^3}{7}-\frac{3 y^6}{49}+O\left(y^9\right) } -\left(1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^{6}\right)
   \right)\right)$$ Long division and simplification lead to
$$A=\frac{9}{14}-\frac{y}{7}-\frac{121 y^2}{1176}+O\left(y^3\right)$$ which shows more than the limit.
A: For positive integers, we have the well known binomial formula
$$(1 + x)^n = \sum_{k\geq 0} \binom{n}{k} x^k = \sum_{k=0}^n \binom{n}{k} x^k$$
where
$$\binom{n}{k} = \frac{(n)_k}{k!} = \frac{n(n - 1)\cdots(n - k + 1)}{k!}$$
$(n)_k$ is the Pochhammer symbol. It is not difficult to show that if we replace $n$ by an arbitrary number $\alpha$, and define analogously
$$\binom{\alpha}{k} =  \frac{\alpha(\alpha - 1)\cdots(\alpha - k + 1)}{k!}$$
we get the Taylor series
$$ (1 + x)^\alpha = \sum_{k\geq 0} \binom{\alpha}{k}x^k = 1 + \alpha x + \mathcal{O}(x^2)$$
Substituting $\frac{1}{x^2} + \mathcal{O}\left(\frac{1}{x^3}\right)$ into the Taylor series for $(1+x)^{\frac{1}{7}}$ yields the desired result.
Edit: Note that
$$\left[\frac{1}{x^2} + \mathcal{O}\left(\frac{1}{x^3}\right)\right]^2 = \mathcal{O}\left(\frac{1}{x^4}\right)$$
so we have
\begin{align*}
\left(1+\frac{1}{x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\right)^\frac{1}{7} &= 1 + \frac{1}{7}\left[\frac{1}{x^2} + \mathcal{O}\left(\frac{1}{x^3}\right)\right] + \mathcal{O}\left(\left[\frac{1}{x^2} + \mathcal{O}\left(\frac{1}{x^3}\right)\right]^2\right)\\
&= 1 + \frac{1}{7x^2} + \mathcal{O}\left(\frac{1}{x^3}\right) + \mathcal{O}\left(\frac{1}{x^4}\right)\\
&=  1 + \frac{1}{7x^2} + \mathcal{O}\left(\frac{1}{x^3}\right)
\end{align*}
A: There is no need for so much work as you have done.
Just note that the expression under limit is of the form $$x^2(A^{1/7}-B)$$ where both $A, B$ tend to $1$. Then we can write the above expression as $$x^2\left(\frac{A^{1/7}-1}{A-1}\cdot(A-1)+1-B\right)$$ Now $x^2(1-B)\to 1/2$ and $$x^2(A-1)=\frac{x^2(x-1)}{x^3+1}\to 1$$ and the desired limit is $$\frac{1}{7}+\frac {1}{2}=\frac {9}{14}$$

The above can be converted into the kind of asymptotics you need. Thus for example you can write $$B=1-\frac{1}{2x^2}+o(x^{-2})$$ and $$A^{1/7}=1+\frac{A-1}{7}+o(x^{-2})$$ and you can get desired answer.
You should also observe that you don't need to multiply/divide/compose Taylor series here. For most typical limit problems that is usually the case.

Even after years of advertising (via my answers on mathse) the limit formula $$\lim_{x\to a} \frac {x^n-a^n} {x-a} =na^{n-1}$$ is not widely used. Perhaps I need to make more effort in this direction.
