# Limit involving L'Hopital's Rule

1. Sep 19, 2012

### Mangoes

1. The problem statement, all variables and given/known data

$\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x - (1 + x)}{x^n}$

where n is a positive integer

3. The attempt at a solution

The numerator approaches 0 as x approaches 0 from the right. The denominator approaches 0 with whichever positive integer, n. This gives an indeterminate form, so I can apply L'Hopital's Rule:

$\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x - 1}{nx^{n-1}}$

Now, here's where I'm not too sure...

The numerator will still approach 0. The denominator will still approach 0 as x approaches 0, it's just being multiplied by some constant. If I apply L'Hopital's Rule again:

$\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x}{n(n-1)x^{n-2}}$

The numerator now reaches 1 as x approaches 0.

I'm unsure of how to evaluate the limit because I'm unsure of how to treat the denominator.

If n was some large positive integer, I'd just get (1/0) as x approaches 0 which would yield to infinity... But if n had been some very small integer, the denominator would have eventually simplified to some constant term and evaluating the limit would not give infinity.

How do I actually go on to evaluate the limit without assigning a value for n?

Would it be legal to just specify the two specific cases (n = 1 and n = 2) where the limit doesn't evaluate towards infinity and then just give the general case for n > 2?

Last edited: Sep 19, 2012
2. Sep 19, 2012

### daveb

Do you know the series expansion for ex? Makes it a lot easier.

(But yes, you would consider each case)

3. Sep 19, 2012

### Mangoes

Nope. Don't know the series expansion for anything... Don't know what that is.

I'm in the first couple weeks of Calc 2.

I just wanted to make sure my thought process was correct throughout. Thanks.

4. Sep 19, 2012

### Staff: Mentor

After two applications of L'Hopital's Rule, the numerator is approaching 1 and the denominator is approaching 0 (but remains positive), so the expression is approaching infinity.

5. Sep 19, 2012

### HallsofIvy

Staff Emeritus
And, as long as n> 1, the denominator goes to 0. What does that tell you?

Do the cases n= 0 and n= 1 separately.

6. Sep 19, 2012

### Mangoes

I'm a little confused now...

Wouldn't n = 0 be omitted since there's a restriction based on n being only positive integers?

From what I understand, n = 1 and n = 2 are the two cases where the expression doesn't grow infinitely large. Any integer larger than 2 will leave (1/0). Shouldn't it be n > 2?

7. Sep 19, 2012

### SammyS

Staff Emeritus
It looks to me that you are correct on all accounts.

So, what do you get for each case: n = 1 and n = 2 ?

8. Sep 19, 2012

### Mangoes

I worked it out and got at n = 1, the limit is 0 and at n = 2, the limit is 1/2

9. Sep 19, 2012

### SammyS

Staff Emeritus
Excellent !