Limit involving L'Hopital's Rule

[Mangoes]

Homework Statement

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

where n is a positive integer

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?

 

[daveb]

Do you know the series expansion for e^x? Makes it a lot easier.

(But yes, you would consider each case)

 

[Mangoes]

Do you know the series expansion for e^x? Makes it a lot easier.

(But yes, you would consider each case)

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.

 

[Mark44]

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.

 

[HallsofIvy]

Homework Statement

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

where n is a positive integer

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.

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.

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?

 

[Mangoes]

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.

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?

 

[SammyS]

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?

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

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

 

[Mangoes]

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

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

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

 

[SammyS]

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

Excellent !