Find $a, b$ so that $\lim_{n \to \infty}((1-n^3)^{\frac{1}{3}}-an-b)=0$ If we go by definition, then we have to find $a,b$, so that for every $\varepsilon >0, \exists  N(\varepsilon), $ so that $\forall n>N(\varepsilon)$
$$|(1-n^3)^{\frac{1}{3}}-an-b| < \varepsilon$$
My attempt,
$$ \left| \frac{1-n^3-an\left(\sqrt[3]{1-n^3}^2\right)-b\left(\sqrt[3]{1-n^3}\right)^2}{\left(\sqrt[3]{1-n^3}\right)^2} \right|< \varepsilon$$
but I'm stuck and I'm not sure if this is how you find $a$ and $b$.

You want $(1-n^3)^{\frac 13} \rightarrow an+b$
If $(1-n^3)^{\frac 13} = an+b + \delta$
Then $(1-n^3) = (an+b + \delta)^3$
$1-n^3 = (an+b + \delta)(a^2n^2+b^2 + \delta^2+2abn+2an \delta + 2b \delta)$
$1-n^3 = a^3n^3+ab^2n + an\delta^2+2a^2bn^2+2a^2n^2 \delta + 2abn \delta$
$+a^2bn^2+b^3 + b\delta^2+2ab^2n+2abn \delta + 2b^2 \delta$
$+a^2n^2\delta+b^2\delta + \delta^3+2abn\delta+2an \delta^2 + 2b \delta^2$
$1-n^3 = a^3n^3+(3a^2b+3a^2 \delta)n^2 +(3ab^2 + 3a\delta^2+ 6ab \delta )n + 3b\delta^2 + 3b^2 \delta+b^3 + \delta^3$
Compare coefficients of $n^3$ to get $a=-1$
$1 = (3b+3 \delta)n^2 -(3b^2 + 3\delta^2+ 6b \delta )n + 3b\delta^2 + 3b^2 \delta+b^3 + \delta^3$
Coefficient of $n^2$ is $3b+3\delta$, where $\delta \rightarrow 0$. Therefore set $b=0$
