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tikki14 · May 11, 2012, 11:39 AM

$\lim_{ x \to \infty} \frac{x^3 + x^2 -1}{e^{2x} + e^x + x^2}$

ebaines · May 13, 2012, 06:36 PM

Originally Posted by tikki14

$\lim_{ x \to \infty} \frac{x^3 + x^2 -1}{e^{2x} + e^x + x^2}$

If you apply l'Hospital's rule 3 times you should be able to determine the limit as x goes to infinity. Post back if you need more clarification.

tikki14 · May 14, 2012, 03:53 AM

Yes, that's right, ebaines. But what operations should I do in order to determine the limit, except using l'Hospital's rule?

For example, this limit can be determined in two ways:

$\lim_{ x \to -2 } \frac{x^2 -4}{x^2 + 5x + 6}$

Method 1 (l'Hospital rule):

$\lim_{x \to -2} \frac{2x}{2x+5} = \frac{-4}{-4+5} =-4$

Method 2:

$\lim_{x to -2} \frac{(x-2)(x+2)}{(x+2)(x+3)} = \lim_{ x \to -2} \frac{x-2}{x+3} = \frac{-4}{1} = -4$

It's the second methos I'm looking for. L'Hospital rule is easy to apply, but I need to sole it both ways.