# Switching Variables in L'Hopital, Series, Limits

My Question is from lines 2 to 3. How did you get from lim x to infinity to limit t approaching zero from the positive?

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You make the substitution, $x=1/t$. – Gerry Myerson Feb 10 '13 at 2:05

The following substitution was used: $\quad\displaystyle \;x = \dfrac 1t.\;$
But this substitution requires that we determine to what value $t$ approaches as $x\to \infty$:
And so we note that, as $\;x \to \infty$, $\;\;x = \dfrac 1t \implies t = \dfrac1x \to 0^+,\;\;$ so we can now rewrite our limit as the limit of a function of $t$: $$\lim_{x\to \infty}\,x\ln\left(1+\frac 2x\right)\;\;=\;\; \lim_{t\to 0^+}\,\frac{\ln (1+2t)}{t}\tag{type \,\frac 00}$$
And so now $L'Hopital is used. - +1$~~~~~~~~~~~\$ – Babak S. Feb 10 '13 at 8:15