Calculate $\lim_{x\to +\infty} x(\sqrt{x^2+1}- \sqrt[3]{x^3+1})$.
First thing came to my mind is to simplify this to something easier. So multiply the numerator and the denominator by something like we used to do when two square roots involves. But I am trying to find that suitable term but it seems out of my reach. Can anybody help me to solve this ? Any hint or help would be nice . Thanks.
write your term as $$\frac{\sqrt{x^2+1}-\sqrt[3]{x^3+1}}{\frac{1}{x}}$$ and use L'Hospital
Note that $$x(\sqrt{x^2+1}-x)\to \frac{1}{2}$$ since
$$x(\sqrt{x^2+1}-x)=\frac{x}{\sqrt{x^2+1}+x}=\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}\to \frac{1}{2}$$
One sees similary that
$$x^2(\sqrt[3]{x^3+1}-x)\to \frac{1}{3}$$ and thus
$$\frac{x^2(\sqrt[3]{x^3+1}-x)}{x}\to 0$$ Thus your limit is $\frac{1}{2}$.
HINT:
As lcm$(2,3)=6$ use $\dfrac{a^6-b^6}{a-b}=\sum_{r=0}^5a^rb^{5-r}$
$\dfrac{(1+h^2)^{1/2}-(1+h^3)^{1/3}}{h^2}=\dfrac{(1+h^2)^3-(1+h^3)^2}{h^2}\cdot\dfrac1{\sum_{r=0}^5(1+h^2)^r(1+h^3)^{5-r}}$
Now $\lim_{h\to0}(1+h^2)^r(1+h^3)^{5-r}=1\cdot1$ for finite $r$
Set $1/x=h\implies h\to0^+$
$$\lim_{h\to0^+}\dfrac{(1+h^2)^{1/2}-(1+h^3)^{1/3}}{h^2}$$
Now use Binomial series $(1+h)^n=1+nh+\dfrac{n(n-1)}2h^2+O(h^3)$
Hint :
Use that $$a-b=\frac{a^6-b^6}{(a^2+ab+b^2)(a^3+b^3)}$$ where $a=\sqrt{x^2+1}$ and $b=\sqrt[3]{x^3+1}$ to have $$\begin{align}\lim_{x\to\infty}x(a-b)&=\lim_{x\to\infty}\frac{(a^6-b^6)x}{(a^2+ab+b^2)(a^3+b^3)}\\\\&=\lim_{x\to\infty}\frac{((x^2+1)^3-(x^3+1)^2)x}{(x^2+1+\sqrt{x^2+1}\sqrt[3]{x^3+1}+(x^3+1)^{2/3})((x^2+1)^{3/2}+x^3+1)}\end{align}$$ Now divide top and bottom by $x^5$.
You have that : $$\begin{align}x(\sqrt{x^2+1}-\sqrt[3]{x^3+1}) &= x^2\left(\sqrt{1+{1\over x^2}}-\sqrt[3]{1+{1\over x^3}}\right)\\ & =x^2\left(1+{1\over 2x^2}-1 + o\left({1\over x^2}\right)\right)\\ &= {1\over 2} + o(1)\end{align}$$
Since $$\sqrt[a]{1+h} = 1+ {h\over a} + o(h)$$ for $a\gt 0$
The idea is that $\sqrt{x^2+1}$ is “the same as $x$ at $\infty$” and similarly for $\sqrt[3]{x^3+1}$.
So a good strategy could be “subtracting and adding $x$”, then considering separately the limits we get: $$ \lim_{x\to\infty}x(\sqrt{x^2+1}-x)-\lim_{x\to\infty}x(\sqrt[3]{x^3+1}-x) $$ Do the first with $x=1/t$ and then, realizing only $t^2$ is involved, $t^2=u$: $$ \lim_{t\to0^+}\frac{\sqrt{1+t^2}-1}{t^2}= \lim_{u\to0^+}\frac{\sqrt{1+u}-1}{u}=\frac{1}{2} $$ because this is the derivative of $u\mapsto\sqrt{1+u}$ at $0$.
For the second, the substitution $x=1/t$ gives $$ \lim_{t\to0^+}\frac{\sqrt[3]{1+t^3}-1}{t^2}= \lim_{t\to0^+}\frac{\sqrt[3]{1+t^3}-1}{t^3}t $$ Now the fraction is the derivative at $0$ of $u\mapsto\sqrt[3]{1+u}$, so the limit is $0$.