Undefined Limit?

1. Jun 7, 2012

dan38

1. The problem statement, all variables and given/known data
Don't understand why the limit (e^x - 1)/x^2 is undefined as x approaches 0?
Can't you use L'hopital's rule to get the value as 1/2?

2. Relevant equations

3. The attempt at a solution

2. Jun 7, 2012

micromass

Staff Emeritus
L'hopitals rule certainly applies here. But I don't see how you use that to get 1/2. Perhaps you can provide your work?

3. Jun 7, 2012

dan38

(e^x - 1) / x^2
Differentiate top and bottom
e^x / 2x
Differentiate top and bottom
e^x / 2
Sub in x = 0
1/2?

4. Jun 7, 2012

algebrat

you applied l'Ho(s)pital's rule the second time, when it was not of the indeterminate form. What is the limit of e^x/(2x)?

5. Jun 7, 2012

Muphrid

I'm not sure why, but I suspect there may be a subtlety here in how l'Hopital's rule can be applied because l'Hopital should be a simple shortcut in lieu of a Taylor series, yet a Taylor series analysis yields (e^x-1)/x^2 -> (x + x^2/2 + ...)/x^2 -> 1/x + 1/2.

6. Jun 7, 2012

dan38

I dont understand algebrat, you can't divide by zero?

7. Jun 7, 2012

Muphrid

He's saying that l'Hopital's rule only applies when you have an indeterminate form: things like $0 \times \infty$ are indeterminate, as is $0/0$, but in the context of limits, $1/0$, while undefined, is not indeterminate. $1/0$ either gives $\pm \infty$ or the limit does not exist, as in this case.

8. Jun 8, 2012

micromass

Staff Emeritus
Indeed, l'Hopitals rule only applies to situations like

$$\frac{0}{0}~\text{and}~\frac{\pm \infty}{\pm \infty}$$

Now the situation is $\frac{1}{0}$, so we can not appy L'Hopitals rule.

9. Jun 8, 2012

dan38

oh i see, thanks!
doing a question: x^2*e^x with x---> infinity
changed it to
x^2/e^(-x)
used L'hopital's rule
2x/-e^(-x)
and again
2/e^(-x)
where do I go from here? :S

10. Jun 8, 2012

Muphrid

You've missed the point. You can't just apply l'Hopital's rule whenever you want. You can only apply it when the functions in the limit would evaluate to something indeterminate. After you applied l'Hopital once, you got, essentially, $\infty/0$. This is not indeterminate. You can't apply l'Hopital again.

Honestly, I would recommend abandoning l'Hopital's rule and just Taylor expanding things. It may require a little extra crunching, but it's a lot more unambiguous and idiot-proof.

11. Jun 8, 2012

Ratch

dan38,

The limit of the term as x-->0 is not defined because it is +∞ when x-->0+ and -∞ when x-->0- . The limit has to be defined in order to use L'Hospital's rule.

Ratch

12. Jun 8, 2012

dan38

2x/-e^(-x)
so how would I go from here then?

13. Jun 8, 2012

clamtrox

Plug in the value x=0.

14. Jun 8, 2012

HallsofIvy

Staff Emeritus
As x goes to infinity, the numerator goes to infinity and e-x goes to 0. micromass had just told you that you cannot use L'Hopitals rule for that. And you don't need to. The numerator of a fraction going to infinity while the denominator goes to 0, the fraction goes to infinity. Actually, that should have been obvious from the start. $x^2e^x$ is the product of two functions both of which go to infinity.

15. Jun 8, 2012

Ratch

dan38,

You don't go anywhere. That function's limit is undefined. It has a one value when x-> is approached from the left, and another value when approached from the right. Isn't that what I said earlier?

Ratch