Finding $a,b$ such that $\lim\limits_{x\to \infty} (\sqrt[3]{1+x^2+x^3}-ax-b)=0$

Question

Given $$ \lim_{x\to \infty} (\sqrt[3]{1+x^2+x^3}-ax-b)=0. $$ What is the value of $a$ and $b$?

Answer 1

\begin{align*} \left(1+x^2+x^3\right)^{\frac{1}{3}} &= x\left(1+\frac{1+x^2}{x^3}\right)^{\frac{1}{3}} \\ &= x\left(1+ \frac{1+x^2}{3x^3} +o(\frac{1}{x})\right) \\ &= x+ \frac{1}{3} + o(1) \\ &\Rightarrow a=1,b=\frac{1}{3} \end{align*}

Answer 2

Original Question:

In fact, \begin{align*} \sqrt{1+x^{2}+x^{3}}-ax-b&>x^{3/2}-ax-b\\ &=x^{3/2}\left(1-\dfrac{a}{x^{1/2}}-\dfrac{b}{x^{3/2}}\right), \end{align*} and that $1-a/x^{1/2}-b/x^{3/2}$ is actually strictly positive for large $x>0$, so the whole term tends to $\infty$.

Answer 3

$$ \left( x + \frac{1}{3} - \frac{1}{9x} \right)^3 \; \; < \; \; x^3 + x^2 + 1 \; \; < \; \; \left( x + \frac{1}{3} \right)^3 $$ for, say, $x \geq 3 \; , \; \;$ as $$ x^3 + x^2 - \frac{5}{27} + \frac{1}{81 x^2} - \frac{1}{729x^3} \; \; < \; \; x^3 + x^2 + 1 \; \; < \; \; x^3 + x^2 + \frac{x}{3} + \frac{1}{27} $$ so with $x \geq 3 \; , \; \;$ we have $$ x + \frac{1}{3} - \frac{1}{9x} \; \; < \; \; \sqrt[3]{ \; \; x^3 + x^2 + 1 \; \; \; \; } \; \; < \; \; x + \frac{1}{3} $$ $$ - \frac{1}{9x} \; \; < \; \; \; \; \sqrt[3]{ \; \; x^3 + x^2 + 1 \; \; \; \; } \; \; - \; \; x \; - \; \frac{1}{3} \; \; \; \; \; < \; \; 0 $$

Answer 4

\begin{align*} \sqrt[3]{1+x^2+x^3}-ax-b&=\frac{(1+x^2+x^3)-(ax+b)^3}{(\sqrt[3]{1+x^2+x^3})^2+(ax+b)\sqrt[3]{1+x^2+x^3}+(ax+b)^2}\\ &=\frac{(1-a^3)x^3+(1-3a^2b)x^2-3ab^2x+(1-b^3)}{(\sqrt[3]{1+x^2+x^3})^2+(ax+b)\sqrt[3]{1+x^2+x^3}+(ax+b)^2}\\ &=\frac{(1-a^3)x+(1-3a^2b)-\frac{3ab^2}{x}+\frac{1-b^3}{x^2}}{(\sqrt[3]{\frac{1}{x^3}+\frac{1}{x}+1})^2+(a+\frac{b}{x})\sqrt[3]{\frac{1}{x^3}+\frac{1}{x}+1}+(a+\frac{b}{x})^2} \end{align*}

As $x\to\infty$, the limit exists if and only if $1-a^3=0$, i.e., $a=1$.

When $a=1$, the limit is $\displaystyle\frac{1-3b}{3}$.

So, $\displaystyle b=\frac{1}{3}$.