# l'hospital

• Sep 24th 2008, 05:25 AM
Linnus
l'hospital
Hey!
How do you do this problem by L'hospital.

$\lim_{x \to \infty} e^x-x^2$

I know you can look at it and say $e^x$ is growing faster than $x^2$.

• Sep 24th 2008, 12:58 PM
mr fantastic
Quote:

Originally Posted by Linnus
Hey!
How do you do this problem by L'hospital.

$\lim_{x \to \infty} e^x-x^2$

I know you can look at it and say $e^x$ is growing faster than $x^2$.

Why use l'Hospital?

Substitute the power series for e^x, simplify and take the limit.
• Sep 24th 2008, 01:38 PM
Linnus
L'hospital because that's what the book wants...

I thought power series was an infinite series...so what would you substitute for e^x?
• Sep 24th 2008, 01:49 PM
mr fantastic
Quote:

Originally Posted by Linnus
L'hospital because that's what the book wants...

I thought power series was an infinite series...so what would you substitute for e^x?

Maclaurin Series -- from Wolfram MathWorld The one for e^x is there somewhere.

I don't care what the book wants, using l'Hopital's rule here is ridiculous.

By the time you figure out how to apply l'Hopital's rule to this problem, you could have solved it three other ways and gone on to answer a dozen other problems.

While I'm warmed up, I'll also say that l'Hopital's rule is the refuge of the lazy and the incompetent nine times out of ten.
• Sep 24th 2008, 01:54 PM
Linnus
I know the series for e^x. I'm just saying since it goes on forever, how do you know when to stop?

Ps: I've figure out another way to solve it by expressing the expressions as ln's.
• Sep 24th 2008, 01:59 PM
mr fantastic
Quote:

Originally Posted by Linnus
I know the series for e^x. I'm just saying since it goes on forever, how do you know when to stop?

[snip]

Since all the terms are positive, I wouldn't have thought this was even an issue when taking the limit x --> +oo ..... Clearly you just need to make sure you use more than the first three terms ....
• Sep 24th 2008, 02:05 PM
Linnus
I'm still a bit confused.

"using more than first 3 terms" is kinda vague. It still doesn't tell me how many terms I should use and why.

Also if we were to change the 2nd term in the original problem to something that is very similar e^x then would the substituting with the series work?
• Sep 24th 2008, 02:18 PM
mr fantastic
Quote:

Originally Posted by Linnus
I'm still a bit confused.

"using more than first 3 terms" is kinda vague. It still doesn't tell me how many terms I should use and why.

Also if we were to change the 2nd term in the original problem to something that is very similar e^x then would the substituting with the series work?

I should have said "at least first three terms".

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \, ....$

What happens if you subtract x^2 from a series using less than three terms ....?