Stuck on Applying L'Hopital's Rule to Indeterminant Form

Click For Summary
SUMMARY

The discussion focuses on applying L'Hopital's Rule to evaluate the limit of the expression (x - (x + 2)e^(1/x)) as x approaches infinity, which presents an indeterminate form of infinity minus infinity. Participants clarify that the limit can be transformed into a suitable form for L'Hopital's Rule by reexpressing it as (1 - e^(1/x))/(1/x). By substituting y = 1/x and taking the limit as y approaches 0, the problem can be resolved effectively.

PREREQUISITES
  • Understanding of limits and indeterminate forms in calculus
  • Familiarity with L'Hopital's Rule and its application
  • Knowledge of exponential functions and their behavior as x approaches infinity
  • Ability to manipulate algebraic expressions to facilitate limit evaluation
NEXT STEPS
  • Study the application of L'Hopital's Rule in various indeterminate forms
  • Explore the behavior of exponential functions as limits approach infinity
  • Practice transforming complex expressions into simpler forms for limit evaluation
  • Learn about alternative methods for evaluating limits, such as Taylor series expansion
USEFUL FOR

Students studying calculus, educators teaching limit concepts, and anyone seeking to master the application of L'Hopital's Rule in solving indeterminate forms.

Redoctober
Messages
46
Reaction score
1

Homework Statement



I know it needs L'Hopital but i can't get it to the indeterminant form

Limit (x-(x+2)e^(1/x)) ((inf - inf ))
x-->inf

I reached until here and then i got stuck

(1-((x+2)e^(1/x))/x)/(1/x) (1 -inf/inf)/0
 
Physics news on Phys.org
Separate limit into [itex]\lim_{x \rightarrow \infty} x(1 - e^{\frac{1}{x}}) - 2e^{\frac{1}{x}}[/itex].

Second term is easy.

First term is indeterminate of the form infinity*0, so reexpress it as [itex]\frac{1 - e^\frac{1}{x}}{\frac{1}{x}}[/itex], then let [itex]y = \frac{1}{x}[/itex] and take the limit as [itex]y \rightarrow 0[/itex] using L' Hopital's Rule.
 
Oh i see :D ! Thanks a lot :)
 

Similar threads

Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
10
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
35
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
1K