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

[RogerDodgr]

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

Homework Statement

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

Homework 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.

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.

 

[Mentallic]

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}&lt;0

\frac{x^{3}+2x-2x^{3}-x^{2}}{2x+1}&lt;0

\frac{-x^{3}-x^{2}+2x}{2x+1}&lt;0

\frac{-x(x+2)(x-1)}{2x+1}&lt;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

 

[statdad]

Since x \to \infty you can assume that x &gt; 1 so that 2x &lt; x^3

then, for such x
<br /> \frac{x^3 + 2x}{2x+1} &lt; \frac{x^3 + x^3}{2x} = \frac{2x^3}{2x} = x^2<br />