Limit involving L'Hopital's Rule

In summary: Now, for n > 2, the limit is easy to evaluate, right?In summary, the limit of the given expression as x approaches 0 from the right is 0 for n = 1, 1/2 for n = 2, and undefined for n > 2.
  • #1
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Homework Statement



[itex]
\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x - (1 + x)}{x^n}
[/itex]

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:

[itex]
\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x - 1}{nx^{n-1}}
[/itex]

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:

[itex]
\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x}{n(n-1)x^{n-2}}
[/itex]

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?
 
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  • #2
Do you know the series expansion for ex? Makes it a lot easier.

(But yes, you would consider each case)
 
  • #3
daveb said:
Do you know the series expansion for ex? 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.
 
  • #4
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
Mangoes said:

Homework Statement



[itex]
\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x - (1 + x)}{x^n}
[/itex]

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:

[itex]
\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x - 1}{nx^{n-1}}
[/itex]

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:

[itex]
\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x}{n(n-1)x^{n-2}}
[/itex]

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?
 
  • #6
HallsofIvy said:
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?
 
  • #7
Mangoes said:
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 ?
 
  • #8
SammyS said:
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
 
  • #9
Mangoes said:
I worked it out and got at n = 1, the limit is 0 and at n = 2, the limit is 1/2
Excellent !
 

1. What is L'Hopital's Rule and when is it used?

L'Hopital's Rule is a mathematical rule used to evaluate limits of indeterminate forms, where both the numerator and denominator approach zero or infinity. It is used when direct substitution of the limit variable results in an undefined expression.

2. How does L'Hopital's Rule work?

L'Hopital's Rule states that if the limit of a function f(x) divided by g(x) is an indeterminate form, then the limit of the derivative of f(x) divided by the derivative of g(x) is equal to the original limit. This process can be repeated until the limit is no longer an indeterminate form.

3. What are the conditions for using L'Hopital's Rule?

The conditions for using L'Hopital's Rule are that the limit must be in an indeterminate form (such as 0/0 or ∞/∞), the function f(x) and g(x) must be differentiable in a neighborhood of the limit point, and the limit of the derivative of f(x) divided by the derivative of g(x) must exist.

4. What are some common mistakes when using L'Hopital's Rule?

One common mistake is forgetting to check if the limit is in an indeterminate form before applying the rule. Another mistake is using the rule when the conditions are not met, such as when the limit is not in an indeterminate form or when the derivatives do not exist.

5. Are there any other methods for evaluating limits of indeterminate forms?

Yes, there are other methods such as using algebraic manipulation, factoring, or trigonometric identities. In some cases, it may also be helpful to graph the function or use a calculator to estimate the limit. However, L'Hopital's Rule is often the most efficient method for evaluating limits of indeterminate forms.

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