Substitution for limits

For real-valued function $f$ we say that:

$\underset{x\rightarrow \infty}{\text{lim}}f(x) = L \iff$ for all $\epsilon > 0$ there exists $B > 0$ such that $x>B \implies |f(x) - L| < \epsilon$.

On the other hand:

$\underset{t\rightarrow 0^+}{\text{lim}}f(\frac{1}{t}) = L^* \iff$ for all $\epsilon > 0$ there exists $\delta > 0$ such that $0 < t <\delta \implies |f(\frac{1}{t}) - L^*|< \epsilon$.

If we have the second version, then by letting $\delta = 1/B$ we have that $0 < t < \frac{1}{B} \implies \frac{1}{t} > B \implies |f(\frac{1}{t})-L^*| < \epsilon$.

If $x = \frac{1}{t}$, we can rewrite the last part as $x > B \implies |f(x) - L^*| < \epsilon$ and we have the first version. By the uniqueness of limits, $L = L^*$.

This proves you can perform this change of variable without any trouble, and the limit will be the same if it exists.