Trick for evaluating limits by substituting in 1/n

  • Context: Undergrad 
  • Thread starter Thread starter cuallito
  • Start date Start date
  • Tags Tags
    Limits
Click For Summary
SUMMARY

The discussion centers on a method for evaluating limits in calculus by substituting \( x \) with \( \frac{1}{n} \). This technique transforms limits such as \( \lim_{x \rightarrow \infty} \) into \( \lim_{n \rightarrow 0} \) or \( \lim_{x \rightarrow 0} \) into \( \lim_{n \rightarrow \infty} \). A specific example provided is the limit \( \lim_{n \to \infty}(1 + \frac{1}{n})^n \), which can be rewritten as \( \lim_{x \to 0^+}(1 + x)^{1/x} \). This substitution is linked to the Heine Theorem, which relates to the behavior of functions at limits.

PREREQUISITES
  • Understanding of basic limit concepts in calculus.
  • Familiarity with the substitution method in limit evaluation.
  • Knowledge of the Heine Theorem and its implications in calculus.
  • Ability to manipulate algebraic expressions involving limits.
NEXT STEPS
  • Study the Heine Theorem and its applications in limit evaluation.
  • Learn about the properties of limits and their transformations.
  • Explore examples of limits involving \( \lim_{n \to \infty} \) and substitutions.
  • Investigate the relationship between limits and derivatives, particularly in the context of dense subsets.
USEFUL FOR

Students of calculus, educators teaching limit concepts, and anyone seeking to deepen their understanding of limit evaluation techniques.

cuallito
Messages
94
Reaction score
1
TL;DR
Trick I vaguely remember from Calc 1
Hi, I remember some sort of method for evaluating limits from Calc 1 that involved substituting in 1/n for x and simplifying. Does that sound familiar to anyone? Sorry I know that's vague, but all I can really remember about it. I can't find it mentioned anywhere in Stewart nor online :/

Thank you...
 
Physics news on Phys.org
Well, that would change a ##\lim_{x \rightarrow \infty}## to a ##\lim_{n \rightarrow 0}## or ##\lim_{x \rightarrow 0}## to ##\lim_{n \rightarrow \infty}##. Perhaps there are cases where that's a useful thing to do, but it's not a trick I recall seeing off the top of my head.
 
cuallito said:
Summary: Trick I vaguely remember from Calc 1

Hi, I remember some sort of method for evaluating limits from Calc 1 that involved substituting in 1/n for x and simplifying. Does that sound familiar to anyone? Sorry I know that's vague, but all I can really remember about it. I can't find it mentioned anywhere in Stewart nor online :/

Thank you...
This substitution appears fairly often in calculus textbooks.
Given the limit ##\lim_{n \to \infty}(1 + \frac 1 n)^n##, you can use the substitution x = 1/n, and work with the new limit ##\lim_{x \to 0^+}(1 + x)^{1/x}## and evaluate the new limit.
 
  • Like
Likes   Reactions: cuallito and jim mcnamara
I believe you are talking about the Heine Theorem.
 
shrub_broom said:
I believe you are talking about the Heine Theorem.
I don't think so.
I am not familiar with the Heine Theorem, but I did find online descriptions of the Heine-Borel Theorem and the Heine-Cantor Theorem, neither of which had anything to do with limits, as far as I could see.

Also, the OP mentioned that his question was about Calc 1, which focuses mostly on limits and differentiation.
 
  • Like
Likes   Reactions: fresh_42
I vaguely remember it used to compute derivatives somewhat similar to what Mark44 wrote , maybe in terms of dense subsets, as in :

##f'(x)=Lim_{ n \rightarrow \infty} [ f(x+1/n)-f(x)] / (1/n) ##

Let me try to remember more details.
 
For every function limit ## lim_{x -> x_0} f(x) = A ##, if there is a series s.t. ##lim_{n -> \infty} a_n = x_0##, then ## lim_{n -> \infty} f(a_n) = A##. By the way, ## lim_{x -> x_0} f(x) = A ## iff ##forall~a_n~s.t.~ lim_{n -> \infty} a_n = x_0, lim_{n -> \infty} f(a_n) = A##. This is what I mentioned.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K