Silly Limit Question: Can We Replace x with t?

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Discussion Overview

The discussion revolves around the conditions under which one can replace the variable \( x \) with \( t \) in the limit expression \( \lim_{x \rightarrow \infty} f(x) \) when \( x \) is defined as a function of \( t \). Participants explore the implications of this substitution, particularly in the context of periodic functions and the behavior of \( f(x) \) as \( x \) approaches infinity.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether it is valid to replace \( \lim_{x \rightarrow \infty} \) with \( \lim_{t \rightarrow \infty} \) under the condition that \( x = a\, t + b\, g(t) \) and \( g(t) \) is periodic.
  • Another participant argues against the substitution, providing the example \( x = e^{-t} \) to illustrate that \( \lim_{t\rightarrow\infty} f(x) = f(0) \), assuming \( f(0) \) exists.
  • Some participants note that the validity of the substitution depends on the form of \( g(t) \), suggesting that specific functions like \( g(t) = \tan t \) could affect the outcome.
  • Further clarification is provided that \( g(t) \) should be bounded, with one participant stating that \( |g(t)| < \infty \) is necessary, using \( g(t) = \sin(t) \) as an example.
  • Another participant emphasizes that \( a \) must be a positive real number to ensure that \( x \) approaches positive infinity, and that the boundedness of \( g(t) \) is sufficient without requiring periodicity.
  • It is mentioned that if \( g(t) \) is bounded below by a constant \( M \), then the expression \( at + bg \) will also approach positive infinity as \( t \) increases.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of substituting \( x \) with \( t \) in the limit expression. Some argue that it is not valid without specific conditions on \( g(t) \), while others suggest that certain bounded functions could allow for the substitution. The discussion remains unresolved with multiple competing views.

Contextual Notes

Participants highlight the importance of the conditions on \( a \) and \( g(t) \), noting that the nature of \( g(t) \) can significantly influence the behavior of the limit. There is an emphasis on the need for \( a \) to be positive and for \( g(t) \) to be bounded, but the exact requirements remain a point of contention.

epimorphic
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I have a rather silly limit question.

Consider
\begin{equation}
\lim_{x \rightarrow \infty} f(x)
\end{equation}
and assume it exists. Suppose now that
\begin{equation}
x = a\, t + b\, g(t),
\end{equation}
where [itex]a[/itex] and [itex]b[/itex] are constants and [itex]g(t)[/itex] is a periodic function of [itex]t[/itex]. Now, is it correct to simply replace [itex]\lim_{x \rightarrow \infty}[/itex] by [itex]\lim_{t \rightarrow \infty}[/itex] as [itex]x \rightarrow \infty[/itex] if and only if [itex]t \rightarrow \infty[/itex]? That is, is it correct to write
\begin{equation}
\lim_{x \rightarrow \infty} f(x) = \lim_{t \rightarrow \infty} f(x(t))\;?
\end{equation}
 
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No. Consider [itex]x=e^{-t}[/itex]. [itex]\lim_{t\rightarrow\infty}f(x)=f(0)[/itex], assuming that f(0) exists.
 
It depends on the form of g(t). Think of g(t)=tan t.
 
dalcde said:
No. Consider [itex]x=e^{-t}[/itex]. [itex]\lim_{t\rightarrow\infty}f(x)=f(0)[/itex], assuming that f(0) exists.
This is not the same [type of] question I have asked.

Useful nucleus said:
It depends on the form of g(t). Think of g(t)=tan t.
I should have stated this more explicitly: [itex]\left|g(t)\right| < \infty[/itex], say [itex]\left|g(t)\right| = \sin(t)[/itex], continuous, smooth and infinitely differentiable.
 
epimorphic said:
This is not the same [type of] question I have asked.

Sorry. Didn't read that.
 
Since the second term in the right hand side is always finite, then your assertion is correct.
 
One minor correction: 'a' must be a positive real number. Otherwise 'x' will go to negative infinity, or simply be bounded and periodic.

Also, this should hold for any bounded periodic function, be it infinitely differentiable, not differentiable at all, or even not continuous anywhere. In fact, it actually need only be bounded below, not above.

EDIT: *Any bounded function at all. There is no particular reason why it needs to be periodic.
 
This is because if g is bounded below by M, then bg is bounded below by bM. Since a is positive, [itex]at+bg \geq at+bM[/itex], and at+bM goes to positive infinity.
 
@Useful nucleus and alexfloo: Thanks!

You are are right, "a" has to be a positive real number and yes the only requirement on g(t) should be as you have stated.
 

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