Limit of $f(x)$ and $f(1/x)$? 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.
 A: 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 $N$.
A: Your whole argument is completely filled with vague hand wavy statements, and as always in mathematics, hand-waving to solve a problem in a subject you are not well versed in, this leads to incoherent and incorrect arguments.
So. Slow down. Take five deep breaths. Five more. OK, now promise you will not make any statement you don't know how to justify mathematically, or any statement that is not perfectly clear what it means.
Good, now you're ready to do mathematics.

First of all, notice that you have a statment that is an "if and only if" statement, i.e. $A\iff B$. This means the usual way of proving it is 


*

*Prove $A\implies B$

*Prove $B\implies A$.


So, let's first go to point (1).
What we want to prove is:

If $\lim_{x\to\infty} f(x)=L$, then $\lim_{x\to 0^+} f(\frac1x)=L$

OK, so there are two statements, and we want to prove one from the other. How can we do that? We assume that the first statement is true, and prove, from that, that the second is true. 
So we are now here:
What you assume is true:

$\lim_{c\to\infty} f(x)=L$

What you want to prove:

$\lim_{x\to 0^+} f(\frac1x)=L$


Now what? Well, the statements are fairly basic, so there's really only one thing to do, and if you only learn one thing today, then here is the lesson:
WRITE DOWN THE DEFINITIONS.
Let's take it nice and slow.
What you assume is true:

For all $\epsilon > 0$, there exists some $M\in\mathbb R$ such that if $y>M$, then $|f(y)-L|<\epsilon$.

(I just replaced $x$ with $y$ so it will be easier to understand what is happening later - the statement doesn't change if we rename a variable)
What you want to prove:

For all $\epsilon > 0$, there exists some $\delta>0$ such that if $0<x<\delta$, then $\left|f(\frac1x)-L\right|<\epsilon$

OK, now we're getting somewhere. What we want to prove is a statement of the form "for all $\epsilon>0$, somethingsomething". So, there's only one thing to do, and that's to write a very familiar statement.
Let $\epsilon>0$.
Now, what we need is to find a $\delta$ such that $\left|f(\frac1x)-L\right|$ will be small if a positive $x$ is smaller than $\delta$.
What we know from the assumptions is that there exists some $M$ such that if $y>M$, then $|f(y)-L|$ will be small. 
Now we look at what we have for a while, and we notice the two expressions are very similar. So we get an idea. What if $y=\frac1x$? Hmm, that sounds like a good idea. OK, let's see. If $y=\frac1x$, then as long as $y<M$, we're good. But we can only control $y$ by limiting $0<x<\delta$! Hmm. Ok. We want $$y>M$$ but $y=\frac1x$. So that's the same as $$\frac1x>M$$
but we can safely assume that $M>0$ without loss of generality, so we can divide by $M$ and multiply by $x>0$ to get $$1>Mx\\ x<\frac1M$$
Oh wow! So what we have now is:


*

*If $x<\frac1M$, then $\frac1x > M$

*If $y=\frac1x > M$, then $|f(y)-L|<\epsilon$


So we are more or less done! Putting that together leads us to conclude that if we set $\delta=\frac1M$, then assuming $\frac1x<M$, we can conclude that $\left|f(\frac1x)-L\right|<\epsilon$, which means that $$\lim_{x\to 0^+} f(\frac1x)=L$$
And this direction is DONE!

I will leave it to you to prove $B\implies A$ (i.e., the other way). Just take it slow, and follow the definitions.
A: This is done easily with sequences. Let $x_n$ be a nonzero sequence such that $x_n\to \infty$ as $n\to \infty$. You know then that $f(x_n) \to L$. Furthermore you know that $\frac{1}{x_n}\to 0$ in a decreasing manner since after some $N$ we have $x_n$ is strictly positive. So:
\begin{align} 
\lim_{x\to 0^+} f\left(\frac 1 x\right)  = \lim_{n\to\infty} f\left(\frac{1} {1/x_n}\right) = \lim_{n\to \infty} f(x_n) =L
\end{align} 
Since $x_n$ was an arbitrary sequence. That is one way. The other way just needs to reverse these statements. And you are done! 
