Limit of (-x^(k+1))/e^x: Solving Step by Step with l'Hopital's Rule

In summary: This is how I'd do it, as I tend to think l'Hopital is a rather unintuitive power-tool that should be used as a last resort. At a minimum, I'd change it to ##0 \leq \frac{x^{k+1}}{e^x} \leq ...##though, otherwise that strict inequality would seem to cause problems, as in the limit we have ## 0 \lt 0##, which doesn't seem right.
  • #1
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Homework Statement


I'm trying to do this limit based on a previous thread ( https://www.physicsforums.com/threads/proving-n-x-n-e-x-integrated-from-0-to-infinity.641947/#_=_ )

I got up to the last part of thread where I need to find the limit of:
limit as x approaches infinity of: (-x^(k+1))/e^x

Homework Equations

The Attempt at a Solution


I know that this limit somehow must equal to zero in order to get the right answer, but I did l'Hopital's rule 4 times and it just keeps on going to infinity.

I attached the working out of the whole problem

Really appreciate it if someone could help
 

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  • #2
If you keep using L'Hospitals rule with a polynomial in the numerator and an exponential in the denominator, the numerator's degree will eventually become 0 while the exponential remains in the denominator.
 
  • #3
LCKurtz said:
If you keep using L'Hospitals rule with a polynomial in the numerator and an exponential in the denominator, the numerator's degree will eventually become 0 while the exponential remains in the denominator.

I can't seem to get it to work here because the exponent of the polynomail has a degree k. If the degree of the polynomail is a variable constant I am not sure how i can get it to zero
 
  • #4
ckyborg4 said:
I can't seem to get it to work here because the exponent of the polynomail has a degree k. If the degree of the polynomail is a variable constant I am not sure how i can get it to zero

What's the derivative of any constant?
 
  • #5
ckyborg4 said:

Homework Statement


I'm trying to do this limit based on a previous thread ( https://www.physicsforums.com/threads/proving-n-x-n-e-x-integrated-from-0-to-infinity.641947/#_=_ )

I got up to the last part of thread where I need to find the limit of:
limit as x approaches infinity of: (-x^(k+1))/e^x

Homework Equations

The Attempt at a Solution


I know that this limit somehow must equal to zero in order to get the right answer, but I did l'Hopital's rule 4 times and it just keeps on going to infinity.

Keep going. You must apply l'Hopital [itex]k + 1[/itex] times in total before you get a constant in the numerator.

Alternatively, as every term in the series [itex]e^x = \sum_{n=0}^\infty \frac{x^n}{n!}[/itex] is strictly positive when [itex]x > 0[/itex], we have [itex]e^x > \frac{x^{k+2}}{(k+2)!}[/itex] and hence [tex]0 < \frac{x^{k+1}}{e^x} < \frac{(k+2)!x^{k+1}}{x^{k+2}} = \frac{(k+2)!}{x}.[/tex] Now use the squeeze theorem.
 
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  • #6
pasmith said:
Keep going. You must apply l'Hopital [itex]k + 1[/itex] times in total before you get a constant in the numerator.

Alternatively, as every term in the series [itex]e^x = \sum_{n=0}^\infty \frac{x^n}{n!}[/itex] is strictly positive when [itex]x > 0[/itex], we have [itex]e^x > \frac{x^{k+2}}{(k+2)!}[/itex] and hence [tex]0 < \frac{x^{k+1}}{e^x} < \frac{(k+2)!x^{k+1}}{x^{k+2}} = \frac{(k+2)!}{x}.[/tex] Now use the squeeze theorem.
This is how I'd do it, as I tend to think l'Hopital is a rather unintuitive power-tool that should be used as a last resort.

At a minimum, I'd change it to

##0 \leq \frac{x^{k+1}}{e^x} \leq ...##

though, otherwise that strict inequality would seem to cause problems, as in the limit we have ## 0 \lt 0##
 

1. What is a limit in mathematics?

A limit in mathematics is a fundamental concept that describes the behavior of a function as the input approaches a certain value. It is used to determine the value that a function "approaches" as the input gets closer and closer to a particular value, without ever actually reaching it.

2. How do you solve a limit?

A limit can be solved by evaluating the function at values closer and closer to the limit point and observing the trend of the outputs. This process is called "taking the limit" and can be done algebraically, graphically, or using numerical methods.

3. What are the different types of limits?

There are several types of limits in mathematics, including one-sided limits, infinite limits, and limits at infinity. One-sided limits only consider the behavior of a function as the input approaches a certain value from one side, while infinite limits describe the behavior of a function as the input approaches infinity or negative infinity. Limits at infinity also describe the behavior of a function as the input gets larger and larger in either the positive or negative direction.

4. What is the importance of limits in mathematics?

Limits play a crucial role in calculus and other areas of mathematics, as they allow us to understand the behavior of functions and make predictions about their values. They are also used to define important concepts such as derivatives and integrals.

5. What are some common techniques for solving limits?

Some common techniques for solving limits include using algebraic manipulation, factoring, rationalization, substitution, and L'Hopital's rule. Graphing the function can also provide insight into its behavior and help with solving limits.

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