Silly limit question

  • Thread starter epimorphic
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  • #1
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}
 

Answers and Replies

  • #2
166
0
No. Consider [itex]x=e^{-t}[/itex]. [itex]\lim_{t\rightarrow\infty}f(x)=f(0)[/itex], assuming that f(0) exists.
 
  • #3
It depends on the form of g(t). Think of g(t)=tan t.
 
  • #4
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.

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.
 
  • #5
166
0
This is not the same [type of] question I have asked.
Sorry. Didn't read that.
 
  • #6
Since the second term in the right hand side is always finite, then your assertion is correct.
 
  • #7
192
0
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.
 
  • #8
192
0
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.
 
  • #9
@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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