Proving the limit of a fraction at infinity by replacing the fraction by a larger fraction.

Question

To prove $$\lim_{x \to -\infty} \frac{x^3}{5x^3 - 2x^2 - 4} = \frac{1}{5},$$ one can identify an $N$ such that $x < N$ implies that $$\left|\frac{x^3}{5x^3 - 2x^2 - 4} - \frac{1}{5}\right| < \epsilon$$ by working backwards. That is $$\left|\frac{x^3}{5x^3 - 2x^2 - 4} - \frac{1}{5}\right| = \left|\frac{2x^2 + 4}{5x^3 - 2x^2 - 4}\right|.$$ It would be nice to simplify this rather messy expression; something you can do as long as you do not introduce changes that prevent the final inequality from holding. In this case, the $4$ term in the numerator of $\frac{2x^2 + 4}{5x^3 - 2x^2 - 4}$ is an inconvenience, and it would be nice to remove it. Simply removing this positive term would make the absolute value of the fraction smaller when what is needed is to make the fraction larger. The same thing happens when the expression $-2x^2 - 4$ is removed from the denominator of our fraction. How should I replace this fraction with a simpler fraction that is clearly larger?

Answer 1

Hint: You want $|\frac {\frac 2 x+\frac4 {x^{3}}} {5-\frac 2 x+\frac 4 {x^{3}}}| <\epsilon$. Use the fact that $|5-\frac 2 x+\frac 4 {x^{3}}|>5-|\frac 2 x|-|\frac 4 {x^{3}}|$. I will leave the rest to you.

Answer 2

Assume that $x<-2$, then $|2x^2+4|\le 2x^2+|x|^2= 3x^2$ and $|2x^2+4|\ge 2x^2-4\ge x^2$. Hence your fraction is less than $|3x^2/(5x^3+x^2)|=3/|5x+2|$ if $x<-2$.

Answer 3

I just continue where you stopped and I assume $|x|\geq 2$:

\begin{eqnarray*}\left|\frac{2x^2 + 4}{5x^3 - 2x^2 - 4}\right| & \stackrel{||a|-|b||\leq |a-b|}{\leq} & \frac{2x^2 + 4}{|5|x|^3 -|2x^2 + 4| |}\\ & \stackrel{2x^2+4\leq 3x^2 \leq 3|x|^3}{\leq} & \frac{3x^2}{5|x|^3 - 3|x|^3} \\ & = & \frac 3{2|x|} \end{eqnarray*}