# Show (x^3+2x)/(2x+1) < x^2 as x -> infinite

1. Sep 29, 2008

### RogerDodgr

Show (x^3+2x)/(2x+1) < x^2 as x --> infinite

1. The problem statement, all variables and given/known data
Show (x^3+2x)/(2x+1) < x^2 as x --> infinite

2. Relevant equations
This is not a formal proof. I just need to reduce the left side of the inequality to the point it is fairly obvious that it is less than x^2 as x-->infinite.

3. The attempt at a solution

so far I turned it into
(x^3-1)/(2x+1) + 1 but it is still not broken down enough to be obvious. I'm not sure what to do next.

2. Sep 30, 2008

### Mentallic

Re: Show (x^3+2x)/(2x+1) < x^2 as x --> infinite

I'm not sure how "obvious" you want it to be, but this should do if it isn't going to be a formal proof -

$$\frac{x^{3}+2x}{2x+1}-x^{2}<0$$

$$\frac{x^{3}+2x-2x^{3}-x^{2}}{2x+1}<0$$

$$\frac{-x^{3}-x^{2}+2x}{2x+1}<0$$

$$\frac{-x(x+2)(x-1)}{2x+1}<0$$

The polynomial of x with degree 3 in the numerator dominates the linear polynomial in the denominator, so from here it can be seen that as $$x\rightarrow\infty$$, the value of the fraction $$\rightarrow\infty$$ and since the highest degree of x is negative, the value will approach $$-\infty$$ which is < 0

3. Sep 30, 2008

Since $$x \to \infty$$ you can assume that $$x > 1$$ so that $$2x < x^3$$
then, for such $$x$$
$$\frac{x^3 + 2x}{2x+1} < \frac{x^3 + x^3}{2x} = \frac{2x^3}{2x} = x^2$$