equivalence of 2 limits I want to show that $\lim_{x\rightarrow\infty} f(x) = \lim_{t\rightarrow 0^+}f(1/t)$.
I need to show that the LHS has a limit $L$ if and only if the RHS has limit $L$.
So let's assume that 
$
\lim_{x\rightarrow\infty} f(x) = L 
$, which is equivalent to saying that
$$
\forall \epsilon\ \exists M > 0\ \forall x, x > M\colon |f(x) - L| < \epsilon. \qquad (1)
$$
What I'm trying to derive from the above is
$$
\forall \epsilon\ \exists \delta > 0\ \forall t, t \in (0,\delta)\colon |f(1/t) - L| < \epsilon. \qquad (2)
$$
Now I'm trying to substitute $x$ by $1/t$ in (1), yielding
$$
\forall \epsilon\ \exists M > 0\ \forall t, 1/t > M\colon |f(1/t) - L| < \epsilon,
$$
and thus
$$
\forall \epsilon\ \exists M > 0\ \forall t, 0 < t < 1/M\colon |f(1/t) - L| < \epsilon.
$$
If I replace $M$ by $1/\delta$, then I seem to be getting 
$$
\forall \epsilon\ \exists (1/\delta) > 0\ \forall t, t \in (0,\delta)\colon |f(1/t) - L| < \epsilon.
$$
Is this equivalent to (2)? If so, why?

You seem to have the right idea, but I'm not sure why you wrote $x$ instead of $1/t$ on the last line. You are correct to take $1/M=\delta$, and at that point you are done, are you not? You've shown that for any $\epsilon >0$ there exists $\delta=1/M>0$ such that whenever $t\in(0,\delta)$, $|f(1/t)-L|<\epsilon$. 
It may be worthwhile to note that, when you substitute $x=1/t$, you can assume $t>0$ since $x$ is positive (I guess it's obvious, but the conclusion is that the limit comes from one side, and maybe it should be pointed out why that is).
