# Silly limit question

1. Sep 8, 2011

### epimorphic

I have a rather silly limit question.

Consider

\lim_{x \rightarrow \infty} f(x)

and assume it exists. Suppose now that

x = a\, t + b\, g(t),

where $a$ and $b$ are constants and $g(t)$ is a periodic function of $t$. Now, is it correct to simply replace $\lim_{x \rightarrow \infty}$ by $\lim_{t \rightarrow \infty}$ as $x \rightarrow \infty$ if and only if $t \rightarrow \infty$? That is, is it correct to write

\lim_{x \rightarrow \infty} f(x) = \lim_{t \rightarrow \infty} f(x(t))\;?

2. Sep 8, 2011

### dalcde

No. Consider $x=e^{-t}$. $\lim_{t\rightarrow\infty}f(x)=f(0)$, assuming that f(0) exists.

3. Sep 8, 2011

### Useful nucleus

It depends on the form of g(t). Think of g(t)=tan t.

4. Sep 8, 2011

### epimorphic

This is not the same [type of] question I have asked.

I should have stated this more explicitly: $\left|g(t)\right| < \infty$, say $\left|g(t)\right| = \sin(t)$, continuous, smooth and infinitely differentiable.

5. Sep 8, 2011

6. Sep 8, 2011

### Useful nucleus

Since the second term in the right hand side is always finite, then your assertion is correct.

7. Sep 8, 2011

### alexfloo

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. Sep 8, 2011

### alexfloo

This is because if g is bounded below by M, then bg is bounded below by bM. Since a is positive, $at+bg \geq at+bM$, and at+bM goes to positive infinity.

9. Sep 8, 2011

### epimorphic

@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.