# When to use L'Hospital's Rule...

• Oct 31st 2007, 06:45 AM
Esiuol
When to use L'Hospital's Rule...
I'm having some issues determining when I can and cannot use L'Hospital's Rule. Tips? Suggested approaches?
• Oct 31st 2007, 07:37 AM
Krizalid
You can use it when you have the following typical cases:

#1 Indetermination of the form $\displaystyle \frac00.$

#2 Indetermination of the form $\displaystyle \frac\infty\infty.$

Of course, if you're solving a problem and this one says that you cannot apply the Rule, don't do it :D:D

Many limits can be computed without the Rule, that makes them more interesting.
• Oct 31st 2007, 09:17 AM
Linnus
Just to add to his comment, there are sometime that you manipulate the problem to get to 0/0 or ∞/∞
Such cases are when the limits become one of those forms : 0^0 1^∞ ∞ - ∞, 0⋅∞ ∞^0
• Oct 31st 2007, 09:37 AM
Esiuol
That's what I'm having issues with -- problems where it isn't apparent that l'hospital's rule applies. My professor just told me to practice, but I'm not getting anywhere...I feel like I'm missing something, (probably extremely simple), but we glossed over the material so fast, I'm not surprised I'm having a little difficulty.
• Oct 31st 2007, 09:44 AM
Linnus
when you plug x in and you get any of the form I listed above, usually there is a way to manipluated it to 0/0 or $\displaystyle \frac\infty\infty$

then you can use the L'Hospital's rule
• Oct 31st 2007, 10:18 AM
Esiuol
For example, the limit as t approaches 0 of (e^t)-1/t^3

It looks like "0/0" so I use l'hopital's rule and get (e^t)/3t^2

Looking at my calculator, I know it's infinity, but I don't understand how we get that from using the rule. I get 1/0 then...
• Oct 31st 2007, 10:24 AM
Krizalid
People sometimes apply the Rule thousand of times, but always you need to check if you got indetermination.

So, when you evaluate directly $\displaystyle \frac{e^t}{3t^2},$ there's no indetermination, and the conclusion follows.
• Oct 31st 2007, 10:31 AM
Linnus
Quote:

Originally Posted by Esiuol
For example, the limit as t approaches 0 of (e^t)-1/t^3

It looks like "0/0" so I use l'hopital's rule and get (e^t)/3t^2

Looking at my calculator, I know it's infinity, but I don't understand how we get that from using the rule. I get 1/0 then...

hmm you probably forgot this...
Lim as t-->0 for (e^t)/3t^2 = infinity because at as the denominator goes to zero, the number gets bigger and bigger...
So yea, the L'hospital rule did work for that problem
• Dec 7th 2007, 07:17 PM
angel.white
Can I also use L'Hospital's rule in:
a) $\displaystyle \frac{-\infty}{\infty}$

b) $\displaystyle \frac{\infty}{-\infty}$

c) $\displaystyle \frac{-\infty}{-\infty}$

d) $\displaystyle \frac{-0}{0}$

e) $\displaystyle \frac{0}{-0}$

f) $\displaystyle \frac{-0}{-0}$

(added the zeros, figure they all take the form 0/0, but may as well make sure :))
• Dec 8th 2007, 12:45 AM
Yes because:

Lim (-f(x)) = -Lim(f(x))
and$\displaystyle \frac {d}{dx} (-f(x))$ = $\displaystyle -\frac{d}{dx}(f(x))$

Quote:

Such cases are when the limits become one of those forms : 0^0 1^∞ ∞ - ∞, 0⋅∞ ∞^0
each of these cases have general methods which you would do well to memorise:
$\displaystyle 0^0$:
$\displaystyle lim 0^0 = e^{lim log(0^0)}$
=$\displaystyle e^{lim 0log0}$
then use the method for 0⋅∞

$\displaystyle 1^\infty$:
lim $\displaystyle 1^\infty$ =$\displaystyle e^{lim log(1^{\infty})}$
= $\displaystyle e^{lim \infty log1}$
then use the method for 0⋅∞

$\displaystyle \infty^0$:
$\displaystyle lim \infty^0$ = $\displaystyle e^{lim log (\infty^0)}$
=$\displaystyle e^{lim 0log(\infty)}$
then use method for 0⋅∞

0⋅∞:
Do either of the following:
0⋅∞ = $\displaystyle \frac{0}{1/\infty}$
=0/0
or
0⋅∞ = $\displaystyle \frac {\infty}{1/0}$
=$\displaystyle \infty/\infty$

$\displaystyle \infty - \infty$:
$\displaystyle lim \infty-\infty = log(lim e^{\infty-\infty})$$\displaystyle =log (lim \frac{e^\infty}{e^\infty}) \displaystyle =log (lim \infty/\infty) • Dec 8th 2007, 08:35 AM angel.white Quote: Originally Posted by badgerigar Yes because: Lim (-f(x)) = -Lim(f(x)) and\displaystyle \frac {d}{dx} (-f(x)) = \displaystyle -\frac{d}{dx}(f(x)) each of these cases have general methods which you would do well to memorise: \displaystyle 0^0: \displaystyle lim 0^0 = e^{lim log(0^0)} =\displaystyle e^{lim 0log0} then use the method for 0⋅∞ \displaystyle 1^\infty: lim \displaystyle 1^\infty =\displaystyle e^{lim log(1^{\infty})} = \displaystyle e^{lim \infty log1} then use the method for 0⋅∞ \displaystyle \infty^0: \displaystyle lim \infty^0 = \displaystyle e^{lim log (\infty^0)} =\displaystyle e^{lim 0log(\infty)} then use method for 0⋅∞ 0⋅∞: Do either of the following: 0⋅∞ = \displaystyle \frac{0}{1/\infty} =0/0 or 0⋅∞ = \displaystyle \frac {\infty}{1/0} =\displaystyle \infty/\infty \displaystyle \infty - \infty: \displaystyle lim \infty-\infty = log(lim e^{\infty-\infty})$$\displaystyle =log (lim \frac{e^\infty}{e^\infty})$
$\displaystyle =log (lim \infty/\infty)$

I see, my book was so vague about what cases I could use it, I remember several times that I got it into usable form, but I didn't think it was usable and kept manipulating it until I got something crazy that would work. And when I finished I would look at it bewildered and go "I really hope that's not on the test"