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

Feb 14, 2017

#### Explanation:

The initial form is $\infty \cdot \left(\infty - \infty\right)$

Rewriting

$\ln \left({x}^{2} - x + 1\right) - \ln \left({x}^{2} + x + 1\right) = \ln \left(\frac{{x}^{2} - x + 1}{{x}^{2} + x + 1}\right)$

lets us see that, since $\ln$ is continuous and the limit of the quotient is $1$, we can write this as the indeterminate $\infty \cdot 0$.

So rewrite the original expression as

$\frac{\ln \left({x}^{2} - x + 1\right) - \ln \left({x}^{2} + x + 1\right)}{\frac{1}{2 x + 3}}$.

The initial form of this limit is $\frac{0}{0}$, so we can apply l'Hospital's rule.

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

$\frac{\frac{2 {x}^{2} - 2}{\left({x}^{2} - x + 1\right) \left({x}^{2} + x + 1\right)}}{\frac{- 2}{2 x + 3} ^ 2} = \frac{{\left(2 x + 3\right)}^{2} \left(2 {x}^{2} - 2\right)}{- 2 \left({x}^{2} - x + 1\right) \left({x}^{2} + x + 1\right)}$.

The limit, as $x \rightarrow \infty$ is $\frac{8}{- 2} = - 4$

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

We'll get

$\left(8 {x}^{4} + \text{ terms of lesser degree")/(-2x^4 +" terms of lesser degree}\right)$

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

$\left(8 + \text{ terms that go to "0)/(-2 +" terms that go to } 0\right)$.

So the limit is $\frac{8}{-} 2 = - 4$.

Another way, a bit more explicit

$\frac{{\left(2 x + 3\right)}^{2} \left(2 {x}^{2} - 2\right)}{- 2 \left({x}^{2} - x + 1\right) \left({x}^{2} + x + 1\right)} = - \frac{1}{2} \cdot \frac{4 {x}^{2} + 12 x + 9}{{x}^{2} - x + 1} \cdot \frac{2 {x}^{2} - 2}{{x}^{2} + x + 1}$

The limit at infinity is $- \frac{1}{2} \cdot 4 \cdot 2 = - 4$