Is there a quick way to compute $\lim_{x\rightarrow \infty} x(\sqrt{x^2+1}-\sqrt[3]{x^3+1})$ without either taking Taylor expansions or the general binomial theorem?
My attempt is via the general binomial theorem. I get the answer 1/2. One could (equivalently) expand the entire expression into a Laurent series with constant term 1/2. I was hoping to find a way to obtain this in an elementary way. I am typing on a phone so cannot write in as much detail as I'd like.
On
Uhm... I guess the following is elementary and viable: let $x>0$, then $$x\left((x^2+1)^{1/2}-(x^3+1)^{1/3}\right)=x\frac{(x^2+1)^3-(x^3+1)^2}{\sum_{t=0}^5 (x^2+1)^{t/2}(x^3+1)^{(5-t)/3}}=\\=\frac{3x^5-2x^4+3x^3}{\sum_{t=0}^5x^5\left(1+x^{-2}\right)^{t/2}\left(1+x^{-3}\right)^{(5-t)/3}}=\frac{3-2x^{-1}+3x^{-2}}{\sum_{t=0}^5\left(1+x^{-2}\right)^{t/2}\left(1+x^{-3}\right)^{(5-t)/3}}$$
And the latter goes to $\frac36$ as $x\to \infty$.
$$\lim_{x\rightarrow +\infty} x(\sqrt{x^2+1}-\sqrt[3]{x^3+1})=\lim_{t\rightarrow0^+}\frac{1}{t}\left(\sqrt{\frac{1}{t^2}+1}-\sqrt[3]{\frac{1}{t^3}+1}\right)=$$ $$=\lim_{t\rightarrow0^+}\frac{\sqrt{1+t^2}-1-\left(\sqrt[3]{1+t^3}-1\right)}{t^2}=$$ $$=\lim_{t\rightarrow0^+}\frac{\frac{t^2}{\sqrt{1+t^2}+1}-\frac{t^3}{\sqrt[3]{I1+t^3)^2}+\sqrt[3]{t^3+1}+1}}{t^2}=\frac{1}{2}$$