How do I solve this limit $\lim _{x\to \infty \:}\left(x\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\right)$?

Question

Please I want to check my answer weither is wrong or right, I found that this limit is equal to 1/3, can you check for me please?

$\lim _{x\to \infty \:}\left(x\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\right)$

Without l'hopital, thanks in advance

Answer 1

HINT:

Set $1/x=h\implies h\to0^+$

$F=\lim _{x\to \infty \:}\left(x\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\right)$ $=\lim_{h\to0^+}\dfrac{(1+h^2)^{1/2}-(1+h^3)^{1/2}}{h^2}$

As lcm$(2,3)=6,$ use $a^6-b^6=(a-b)(a^5+a^4b+\cdots+ab^4+b^5)$ to get

$F=\lim_{h\to0^+}\dfrac{(1+h^2)^3-(1+h^3)^2}{h^2}\cdot\dfrac1{\lim_{h\to0^+}\sum_{r=0}^5((1+h^2)^{1/2})^r\cdot((1+h^3)^{1/3})^{(5-r)}}$

$=\lim_{h\to0^+}\dfrac{1+3h^2+3h^4+h^6-(1+3h^3+h^6)}{h^2}\cdot\dfrac1{\sum_{r=0}^51}=\dfrac36$

Answer 2

By using the definitions of $(\sqrt x)'$and$(\sqrt[3]x)'$at $x=1$

As $y=1/x$, it becomes $$\begin{align} \lim_{y\to0}\frac{(1+y^2)^{1/2}-(1+y^3)^{1/3}}{y^2} &=\lim_{y\to0}\frac{(1+y^2)^{1/2}-1}{y^2}-\lim_{y\to0}y\left(\frac{(1+y^3)^{1/3}-1}{y^3}\right)\\&=1/2-0/3=1/2 \end{align}$$