Question

How do you find the Limit of `(2x+3)[ln(x^2-x+1)-ln(x^2+x+1)]` as x approaches infinity?

Answer 1

Please see below.

Explanation:

The initial form is `oo * (oo-oo)`

Rewriting

`ln(x^2-x+1)-ln(x^2+x+1) = ln((x^2-x+1)/(x^2+x+1))`

lets us see that, since `ln` is continuous and the limit of the quotient is `1`, we can write this as the indeterminate `oo * 0`.

So rewrite the original expression as

`[ln(x^2-x+1)-ln(x^2+x+1)]/(1/(2x+3))`.

The initial form of this limit is `0/0`, so we can apply l'Hospital's rule.

It takes some algebra for the numerator, but the quotient of the derivatives is

`((2x^2-2)/((x^2-x+1)(x^2+x+1)))/((-2)/(2x+3)^2) = ((2x+3)^2(2x^2-2))/(-2(x^2-x+1)(x^2+x+1))`.

The limit, as `xrarroo` is `8/(-2) = -4`

One way to see this is to expand (or describe the expansion of) the polynomials.

We'll get

`(8x^4 +" terms of lesser degree")/(-2x^4 +" terms of lesser degree")`

Dividing numerator and denominator by `x^4` (or factoring and reducing) will get us

`(8 +" terms that go to "0)/(-2 +" terms that go to "0)`.

So the limit is `8/-2 = -4`.

Another way, a bit more explicit

`((2x+3)^2(2x^2-2))/(-2(x^2-x+1)(x^2+x+1)) = -1/2 *(4x^2+12x+9)/(x^2-x+1) * (2x^2-2)/(x^2+x+1)`

The limit at infinity is `-1/2 * 4 * 2 = -4`