Show that $x^3-\lambda x+1=0$ has an approximate root $x=\frac{1}{\lambda}$ when $\lambda$ is positive and large I'm currently trying to figure out a way to show that $\frac{1}{\lambda}$ is an approximate root for
$$x^3-\lambda x +1=0$$
when $\lambda$ is positive and large.
My idea is to rearrange the equation as $x=(\lambda x-1)^\frac{1}{3}$ and use a binomial expansion, but I'm kinda unsure how to approach this question.
 A: Rescale to get a proper perturbation problem. Set $z=λx$, then the equation transforms to
$$
\frac1{λ^3}z^3-z+1=0.
$$
Substitute $\epsilon = \frac1{λ^3}$ then
$$
z=1+ϵz^3.
$$
For $ϵ=0$ the solution is $z=1$, for $ϵ\approx 0$ one can obtain the solution as a perturbation series  $z=1+c_1ϵ+c_2ϵ^2+...$ where
$$
c_1ϵ+c_2ϵ^2+...=ϵ(1+c_1ϵ+c_2ϵ^2+...)^3=ϵ+3c_1ϵ^2+3(c_1^2+c_2)ϵ^3+...
$$
Comparing coefficients one obtains $c_1=1$, $c_2=3$, $c_3=12$, ...

This gives in the original scaling
$$
x=\frac1λ+\frac1{λ^4}+\frac3{λ^7}+\frac{12}{λ^{10}}+...
$$
Up to now we can not be sure if this series converges for any finite $λ$.

Using the intermediate value theorem one can show that 

there is indeed a root between $\frac1λ$ and $\frac1λ+\frac2{λ^4}$ or between $\frac1λ+\frac1{λ^4}$ and $\frac1λ+\frac1{λ^4}+\frac6{λ^7}$.

For the first claim, observe first that $x^3-λx+1=\frac1{λ^3}>0$ at $x=\frac1λ$, and at the other point $x=\frac1λ+\frac2{λ^4}$ the polynomial value is
$$
(\frac1λ+\frac2{λ^4})^3-(1+\frac2{λ^3})+1=\frac6{λ^6}+\frac{12}{λ^9}+\frac8{λ^{12}}-\frac1{λ^3}<-\frac3{8λ^3}<0
$$
for $λ>2$.

Using the Rouché theorem one can more exactly show that 

there is exactly one root of $f(z)=1-z+ϵz^3$ with $|z-1|<2ϵ$. 

On the circle $|z-1|=2ϵ$ the difference to $g(z)=1-z$ is bounded via $$|z^3|\le(1+2ϵ)^3=1+6ϵ+12ϵ^2+8ϵ^3<1+(6+\frac32+\frac18)ϵ<1+8ϵ<2$$ for $ϵ<\frac18$. Thus the condition $|f(z)-g(z)|<2ϵ=|g(z)|$ for Rouché is satisfied and there is exactly one root inside this cirlce. 


That is, for $λ>2$ there is exactly one root of the original equation with $$\left|x-\frac1λ\right|<\frac2{λ^4}.$$



Going one step further, one could also show that 

there is a root of $f(z)=1-z+ϵz^3$ with $|z-1-ϵ|<6ϵ^2$. 

Compare $f$ on the circle $|z-1-ϵ|=6ϵ^2$ with $g(z)=1+ϵ-z$ so that $|f(z)-g(z)|=ϵ|z^3-1|=ϵ|z-1|\,|z^2+z+1|$. Now $$|z-1|=ϵ+|z-1-ϵ|<ϵ+6ϵ^2<\frac32ϵ$$ for $ϵ<\frac1{12}$ and then $$|z^2+z+1|\le 3+|z-1|\,|z+2|<3+\frac32ϵ(3+\frac32ϵ)<\frac72$$
so that indeed
$$
|f(z)-g(z)|<ϵ\cdot\frac32ϵ\cdot\frac72=\frac{21}4ϵ^2<6ϵ^2=|g(z)|.
$$


In conclusion, for $λ>\frac52$ there is exactly one root of the original equation satisfying 
    $$
\left|x-\frac1λ-\frac1{λ^4}\right|<\frac6{λ^7}
$$


A: $x^3$ has zero derivation at origin, so it can be approximated by a constant zero function. Then you have:
$$0 - \lambda x + 1 = 0$$
So the root is approximately $\frac1{\lambda}$.
