# Thread: Finding the limit as x tends to infinity

1. ## Finding the limit as x tends to infinity

The question:
Determine the limiting behaviour of $\displaystyle \frac{ln(x^3+1)}{ln(x^2+1)}$ as x -> infinity.

My attempt:
I noticed that both the numerator and denominator approach infinity, making the limit indeterminate. Thus, I tried using L'Hopitals rule. However, in doing so, I can't seem to get an answer even after several differentiations. It seems to constantly become 0/0. Am I attempting this the right way?

Thanks!

2. Using L'Hospital...

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

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

$\displaystyle = \lim_{x \to \infty}\frac{3x^4 + 3x^2}{2x^4 + 2x}$.

This is still $\displaystyle \frac{\infty}{\infty}$ so use L'Hospital again...

$\displaystyle \lim_{x \to \infty}\frac{3x^4 + 3x^2}{2x^4 + 2x} = \lim_{x \to \infty}\frac{12x^3 + 6x}{8x^3 + 2}$.

This is still $\displaystyle \frac{\infty}{\infty}$ so use L'Hospital again...

$\displaystyle \lim_{x \to \infty}\frac{12x^3 + 6x}{8x^3 + 2} = \lim_{x\to\infty}\frac{36x^2 + 6}{24x^2}$.

This is still $\displaystyle \frac{\infty}{\infty}$ so use L'Hospital again

$\displaystyle \lim_{x \to \infty}\frac{36x^2 + 6}{24x^2} = \lim_{x \to \infty}\frac{72x}{48x}$

$\displaystyle = \lim_{x \to \infty}\frac{3}{2}$

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

3. Ahh darn, looks like I made an error right at the start. Thanks!

Also, you can stop after the first application of L'Hopital, since you can divide by the largest powers to get 3/2. (that's a shortcut we learnt at uni)

4. Originally Posted by Glitch
Ahh darn, looks like I made an error right at the start. Thanks!

Also, you can stop after the first application of L'Hopital, since you can divide by the largest powers to get 3/2. (that's a shortcut we learnt at uni)
is 3/2 a typo? denominator has higher degree polynomial so limit is 0.

5. Prove It had the right answer (3/2), then he changed it for some reason.

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

Unless I'm mistaken. (the textbook states 3/2 as the answer)

7. Gah hang on, I'll fix it. It's well known my algebra is shocking...

8. I assure you mine is worse. :P

9. Originally Posted by Glitch
Prove It had the right answer (3/2), then he changed it for some reason.
Ah, didn't bother checking the algebra, just saw that at the end of the first application of L'Hopital's rule Prove It had a cubic over a quartic (now corrected).