Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

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?

share|cite|improve this question
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.

share|cite|improve this answer
+1 $~~~~~~~~~~~$ – Babak S. Feb 10 '13 at 8:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.