Using the theorem of L'Hospital, calculate:
$\displaystyle \lim_{n\to\infty} \frac{ln(x^2+x+1)}{ln(x^3+2x+1)}$
answer:
$\displaystyle \frac{2}{3}$
I presume you mean
$\displaystyle \lim_{x \to \infty} \frac{ln(x^2 + x + 1)}{ln(x^3 + 2x + 1)}$
This form is of the type $\displaystyle \infty / \infty$ so we can use L'Hopital.
$\displaystyle \lim_{x \to \infty} \frac{ln(x^2 + x + 1)}{ln(x^3 + 2x + 1)}$
$\displaystyle = \lim_{x \to \infty} \frac{\frac{1}{x^2 + x + 1} \cdot (2x + 1)}{\frac{1}{x^3 + 2x + 1} \cdot (3x^2 + 2)}$
$\displaystyle = \lim_{x \to \infty} \frac{(x^3 + 2x + 1)(2x + 1)}{(x^2 + x + 1)(3x^2 + 2)}$
This is still indeterminate, but we can now multiply out the numerator and denominator and are left with a polynomial divided by a polynomial. I'll leave the rest to you.
-Dan