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
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
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?