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I seems pretty obvious to me but how I can mathematically show that: $$\lim_{x\rightarrow0^+} f(x) = \lim_{x\rightarrow\infty} f(1/x)$$

Thanks for your guidance.

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at the RHS, sub $y=1/x$, then $lim_{x\rightarrow \infty}y=0^+$ –  Vincent Tjeng Feb 28 '13 at 14:27
    
$x \to 0 \Leftrightarrow 1/x \to \infty$ –  ᴊ ᴀ s ᴏ ɴ Feb 28 '13 at 14:27
    
As I mentioned, I know and it seems pretty obvious. Is there any formal proof for that? I don't know, something more rigorous ... –  Sam Feb 28 '13 at 14:32

1 Answer 1

up vote 3 down vote accepted

This is a fairly direct result of the definitions of the two sorts of limits. Just for clarity, we'll use $x\to 0+$ and $y\to +\infty$. Not 100% necessary, but it can make steps clearer.

We say $\lim_{x\to 0+} f(x) = A$ if, for every $\varepsilon>0$ there is a $\delta>0$ such that for all $x\in(0,\delta)$, $|f(x)-A|<\varepsilon$.

We say $\lim_{y\to +\infty} g(y) = B$ if, for every $\varepsilon>0$ there is a $N$ such that for all $y>N$, $|g(y)-B|<\varepsilon$.

Now, assume $g(y)=f(1/y)$.

If we know $\lim_{x\to 0+} f(x) = A$ exists, then for any $\varepsilon>0$ find a $\delta>0$, so that $|f(x)-A|<\varepsilon$ for $x\in(0,\delta)$. Let $N=\frac{1}{\delta}$. Now if $y>N$ then $\frac{1}{y}\in (0,\delta)$ so $$|g(y)-A| = |f(1/y)-A| < \varepsilon$$

So we've shown if $\lim_{x\to 0+} f(x) = A$ then $\lim_{y\to +\infty} f(1/y) = A$, too.

On the other hand, if we know $\lim_{y\to +\infty} g(y) = A$, then, given $\varepsilon$ and $N$, we can take $\delta = \frac{1}{N}$ if $N>0$ and $\delta = 1$ if $N\leq 0$.

[Technically, the definition for $\lim_{x\to +\infty}$ doesn't require $N$ to be positive, although if we can find such an $N$, we can always choose instead any number $N'\geq N$ - in particular, $N'=\max(N,1)$.]

Basically, given an $\varepsilon$, there is a $\delta$ if and only if there is an $N$.

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