If $\lim_{x\to0} f(x) = L$ then $\lim_{x\to \infty }f(1/x) = L$ . Is the converse true?

Question

If

$$\lim_{x\to0} f(x) = L \implies \lim_{x\to \infty }f(1/x) = L$$

Is the converse true?

$$\lim_{x\to \infty }f(1/x) = L \implies \lim_{x\to0} f(x) = L$$

Why or why not?

So I think its true if I make $y= 1/x$ then it becomes lim $f(y) = L$ as $y $ approches $0$.

So it basically becomes the same thing right? I cant see why it wouldn't be true.

Answer 1

Note that more in general since

$$\lim_{x\to 0} x=\lim_{y\to \pm \infty}\frac1y =0$$

we have that if $\lim_{x\to 0} f(x)$ exists

$$\lim_{x\to 0} f(x)=\lim_{y\to \pm \infty} f\left(\frac1y\right)$$

since both expressions represent the same limit.

As noticed by Hans Lundmark we need to keep attention to the sign and we have that

$$\lim_{x\to 0} f(x)=L \iff \lim_{y\to \pm \infty} f\left(\frac1y\right)$$

but

$$\lim_{y\to +\infty} f\left(\frac1y\right) \implies \lim_{x\to 0^+} f(x)=L $$

$$\lim_{y\to -\infty} f\left(\frac1y\right) \implies \lim_{x\to 0^-} f(x)=L $$

Answer 2

No, it's false. You can only conclude that $\displaystyle\lim_{x \to 0^+} f(x) = L$, since you don't know anything about what happens as $x \to 0^-$.

(Assuming, of course, that the question concerns functions of a real variable $x$.)