For what value of $k$ does $\lim_{x\to \infty} [(x^\alpha+x^2)^k-x]$ have a limit? ($\alpha$ is a constant and $\alpha \ge 3)$ I tried to use L'Hopital's rule, but the power of the leading term of the numerator is larger than the power of the leading term of the denominator.
 A: We are checking
$$\lim_{x \to \infty} \left[\left(x^\alpha + x^2\right)^k - x\right] \tag{1}\label{eq1}$$
and are given that
$$\alpha \ge 3 \tag{2}\label{eq2}$$
Note that
\begin{align}
\left(x^\alpha + x^2\right)^k - x & = \left[x^\alpha\left(1 + x^{2 - \alpha}\right)\right]^k - x \\
& = x^{k\alpha}\left(1 + x^{2 - \alpha}\right)^k - x \tag{3}\label{eq3}
\end{align}
From \eqref{eq2}, $2 - \alpha < 0$ so the term in the brackets will go to $1$ as $x \to \infty$. Thus, for the entire expression to have any possibility of a finite limit requires that
$$k \alpha = 1 \Rightarrow k = \cfrac{1}{\alpha} \tag{4}\label{eq4}$$
since if $k$ is larger, the value in \eqref{eq3} will go to $\infty$ and, if it's smaller, it'll go to $-\infty$. As such, \eqref{eq3} becomes
\begin{align}
x^{k\alpha}\left(1 + x^{2 - \alpha}\right)^k - x & = x\left(1 + x^{2 - \alpha}\right)^k - x \\
& = x\left[\left(1 + x^{2 - \alpha}\right)^k - 1\right] \tag{5}\label{eq5}
\end{align}
Newton's generalized binomial theorem provides that
$$\left(1 + x^{2 - \alpha}\right)^k = 1 + kx^{2 - \alpha} + O\left(x^{2\left(2 - \alpha\right)}\right) \tag{6}\label{eq6}$$
which then gives
$$x\left[\left(1 + x^{2 - \alpha}\right)^k - 1\right] = kx^{\left(2 - \alpha\right) + 1} + O\left(x^{2\left(2 - \alpha\right) + 1}\right) \tag{7}\label{eq7}$$
Using \eqref{eq2}, we have that
$$\left(2 - \alpha\right) + 1 = 3 - \alpha \le 0 \tag{8}\label{eq8}$$
is a non-positive value. Similarly, the error term's power is
$$2\left(2 - \alpha\right) + 1 = 5 - 2\alpha \lt 0 \tag{9}\label{eq9}$$
so it's always negative and, thus, the error goes to $0$ as $x \to \infty$, so it doesn't contribute anything to the limit.
In summary, we have that for $\alpha = 3$ that the first term's $x$ power is $0$ so it becomes just $k$ and, thus,
$$\lim_{x \to \infty} \left[\left(x^\alpha + x^2\right)^k - x\right] = k = \cfrac{1}{3} \tag{10}\label{eq10}$$
and for $\alpha \gt 3$ we instead have
$$\lim_{x \to \infty} \left[\left(x^\alpha + x^2\right)^k - x\right] = 0 \tag{11}\label{eq11}$$
A: Rewrite $(x^\alpha+x^2)^k-x=x^{2k}\left(x^{\alpha-2}+1\right)^k - x=x\left[x^{2k-1}\left(x^{\alpha-2}+1\right)^k - 1\right].$
Assume we want the existence of a finite limit. 
Necessarily $$\lim\limits_{x \to \infty}x^{2k-1}\left(x^{\alpha-2}+1\right)^k = 1. \tag 1$$ This opens two cases 


*

*$k= {1 \over 2},$ which we exclude as $x^{2k-1}=1$ and $\left(x^{\alpha-2}+1\right)^k \to \infty$

*$k<{1\over 2},$ here we have "$0\cdot\infty$". Write $(1)$ as $$\lim\limits_{x \to \infty}\frac{\left(x^{\alpha-2}+1\right)^k}{x^{1-2k}} = 1. $$ This is only true if $\;k(\alpha-2)=1-2k$ or equivalently $\fbox{k= 1/$\alpha$}.$
This is a necessary condition and we need to check if is also sufficient. Set $k={1\over\alpha},$ then
$$\lim_{x\to \infty} \left[(x^\alpha+x^2)^{1\over\alpha}-x\right]=\lim_{x\to \infty}\frac{(1+x^{2-\alpha})^{1\over\alpha}-1}{1\over x}$$ which is equal to $0$ (l'Hospital's rule). The condition ${k= 1/\alpha}$ is also sufficient.
