Finding the Limit of a Fraction with Exponential Terms

  • Thread starter Thread starter PCSL
  • Start date Start date
  • Tags Tags
    Fraction Limit
Click For Summary

Homework Help Overview

The discussion revolves around finding the limit of a fraction involving exponential terms, specifically the expression \(\lim_{n\rightarrow ∞}\frac{3^n+2*5^n}{2^n+3*5^n}\). Participants are exploring different approaches to evaluate this limit.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • One participant attempts to apply L'Hôpital's rule but expresses uncertainty about its appropriateness for the problem. Another suggests factoring out the dominant term, \(5^n\), from the numerator and denominator as a simpler approach. A separate question arises regarding the application of L'Hôpital's rule to a different limit involving a quotient raised to a power.

Discussion Status

Participants are actively engaging with the problem, with some offering alternative methods and questioning the validity of certain approaches. There is a recognition of the dominant term in the limit, and while one participant indicates they have resolved their confusion, others continue to explore the topic.

Contextual Notes

Some participants mention that the problem is part of an early assignment, suggesting a focus on foundational concepts. There is also a note that L'Hôpital's rule may not be applicable in certain contexts, prompting further discussion on appropriate methods for evaluating limits.

PCSL
Messages
146
Reaction score
0
[tex]\lim_{n\rightarrow ∞}\frac{3^n+2*5^n}{2^n+3*5^n}[/tex]

I tried using l'hospital's rule and got

[tex]\lim_{n\rightarrow ∞}\frac{3^n*ln(3)+2(5^n*ln5)}{2^n*ln(2))+3(5^n*ln5)}[/tex]

I'm not quite sure if that is the right way to approach this. This problem is an early one in the assignment so I assume that it is simple and I am just missing something obvious.

Thank you!
 
Physics news on Phys.org
PCSL said:
[tex]\lim_{n\rightarrow ∞}\frac{3^n+2*5^n}{2^n+3*5^n}[/tex]

I tried using l'hospital's rule and got

[tex]\lim_{n\rightarrow ∞}\frac{3^n*ln(3)+2(5^n*ln5)}{2^n*ln(2))+3(5^n*ln5)}[/tex]

I'm not quite sure if that is the right way to approach this. This problem is an early one in the assignment so I assume that it is simple and I am just missing something obvious.

Thank you!

A much simpler approach is to factor 5n out of all terms in the numerator and denominator. Evaluating the limit is pretty easy after that.
 
Mark44 said:
A much simpler approach is to factor 5n out of all terms in the numerator and denominator. Evaluating the limit is pretty easy after that.

so then I'd have

[tex]\lim_{n\rightarrow ∞} \frac{\frac{3}{5}^n+2}{\frac{2}{5}^n+3}[/tex]

Nevermind, I got it thanks!
 
Yes, that's right. The key idea is to find the dominant term in the numerator or denominator, which turns out to be 5n in this problem.
 
For

[tex]\lim_{n\rightarrow ∞}(\frac{n+1}{n})^n[/tex]

is l'hoptials rules the correct way to approach it?
 
Last edited:
No, since L'Hopital's Rule applies to quotients, and that's not what you have here. (The quotient is raised to a nonconstant power.

The usual approach to this type of problem is:
1) Let y = the expression in the limit. Don't include the limit operation.
2) Take the natural log of both sides to get ln y = ln(expression).
3) Use the properties of logs to simplify the right side
4) Take the limit of both sides.
5) Switch the limit and ln operations to get ln(lim y). I.e., this is the log of what you want.
 
Mark44 said:
No, since L'Hopital's Rule applies to quotients, and that's not what you have here. (The quotient is raised to a nonconstant power.

The usual approach to this type of problem is:
1) Let y = the expression in the limit. Don't include the limit operation.
2) Take the natural log of both sides to get ln y = ln(expression).
3) Use the properties of logs to simplify the right side
4) Take the limit of both sides.
5) Switch the limit and ln operations to get ln(lim y). I.e., this is the log of what you want.

Really appreciate it.
 

Similar threads

Replies
17
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
17
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
15
Views
2K
Replies
14
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K