evaluate the limit when x goes to infinity

Question

Evaluate $\displaystyle{\lim_{x \to \infty} (x^3+6x^2+1)^{\frac13}-(x^2}+x+1)^{\frac12}$

I tried writing it as $$\lim_{x \to \infty}\frac{x^3+6x^2+1}{(x^3+6x^2+1)^{\frac23}}-\frac{x^2+x+1}{(x^2+x+1)^{\frac12}}$$ but I do not know how to continue from this point.

I really appreciate your time and effort. Thank you very much!

Answer 1

Let $t=\frac{1}{x}$. Then

$$(x^3+6x^2+1)^{\frac13}-(x^2+x+1)^{\frac12} = \frac{(t^3+6t+1)^{\frac13}-(t^2+t+1)^{\frac12}}{t} = \frac{f(t)-f(0)}{t-0} $$

where $f(t)=(t^3+6t+1)^{\frac13}-(t^2+t+1)^{\frac12}$. Therefore, if it exists, the limit you are looking for is equal to $\lim_{t\to 0^+}\frac{f(t)-f(0)}{t-0}$. Since $f$ is differentiable at $0$, the limit exists and is equal to $f'(0)$.

Using Taylor series, $f(t)=0+\frac{3}{2}t+...$, so $f'(0)=\frac{3}{2}$.

Answer 2

$$\lim_{x \to \infty} (x^3+6x^2+1)^{\frac13}-(x^2+x+1)^{\frac12}$$ Using $(1+t)^a\sim1+at $ as $t\to0$, you have: $$\sqrt[3]{x^3+6x^2+1}-\sqrt{x^2+x+1}=x\sqrt[3]{1+6/x+1/x^3}-x\sqrt{1+1/x+1/x^2}\sim \\ x(1+2/x)-x(1+1/2x)=2-1/2=3/2$$