theorem of L'Hospital

• Jun 22nd 2008, 11:28 AM
Apprentice123
theorem of L'Hospital
Using the theorem of L'Hospital, calculate:

$\displaystyle \lim_{n\to\infty} \frac{ln(x^2+x+1)}{ln(x^3+2x+1)}$

$\displaystyle \frac{2}{3}$
• Jun 22nd 2008, 11:44 AM
topsquark
Quote:

Originally Posted by Apprentice123
Using the theorem of L'Hospital, calculate:

$\displaystyle \lim_{n\to\infty} \frac{ln(x^2+x+1)}{ln(x^3+2x+1)}$

$\displaystyle \frac{2}{3}$

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
• Jun 22nd 2008, 03:51 PM
Apprentice123
Thanks for helping