Solving Basic Limit Questions: When t=0, is it 0, -1, or Undefined?

  • Thread starter Thread starter complexnumber
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary
SUMMARY

The limit function f(t) defined as f(t) = lim(k→∞) (1 - kt²)/(1 + kt²) evaluates to -1 for t ≠ 0 and 1 for t = 0. The reasoning involves substituting t into the original formula, which simplifies to 1 when t = 0, demonstrating that the limit is not undefined. The discussion clarifies the importance of evaluating the original formula versus taking the limit directly, emphasizing that the original formula should be used when t = 0.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with the concept of continuity
  • Knowledge of rational functions
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of limits in calculus
  • Learn about continuity and its implications for limit evaluation
  • Explore rational functions and their behaviors at specific points
  • Practice problems involving limits with different approaches
USEFUL FOR

Students studying calculus, particularly those grappling with limit concepts, and educators seeking to clarify limit evaluation methods in rational functions.

complexnumber
Messages
61
Reaction score
0

Homework Statement



[tex] \begin{align*}<br /> f(t) = \lim_{k \to \infty} f_k(t) = \lim_{k \to \infty} \frac{1 - kt^2}{1 +<br /> kt^2} = \lim_{k \to \infty} \frac{\frac{1}{k} - t^2}{\frac{1}{k} +<br /> t^2} = \frac{0 - t^2}{0 + t^2} = - \frac{t^2}{t^2}<br /> \end{align*}[/tex]

What is the value of limit function [tex]f[/tex] when [tex]t = 0[/tex]? Is it [tex]0[/tex] or [tex]-1[/tex] or undefined? What is the reasoning behind it?

Does anyone know any good websites or books to catch up on these material?

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
None of the above!

If [itex]t\ne 0[/itex] then the limit is -1, obviously.

If t= 0, go back to the original formula: if t= 0, then
[tex]\frac{1- kt}{1+ kt}= \frac{1- 0}{1+ 0}= \frac{1}{1}= 1[/tex]
which is independent of k. The limit, if t= 0, is 1.
 
HallsofIvy said:
None of the above!

If [itex]t\ne 0[/itex] then the limit is -1, obviously.

If t= 0, go back to the original formula: if t= 0, then
[tex]\frac{1- kt}{1+ kt}= \frac{1- 0}{1+ 0}= \frac{1}{1}= 1[/tex]
which is independent of k. The limit, if t= 0, is 1.

Thanks for your reply. I have one question about getting to the solution.

When should I use the original formula first and when should I take the limit first?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 16 ·
Replies
16
Views
3K