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

1 Answer
Feb 14, 2017

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#