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

Today at school we have to determine this limits value, but when the teacher tried, he said maybe you can't determine this limits value without using the Hospital theorem, but try to calculate it without.

$$ \lim_{x\to +\infty} \dfrac{\ln(x)}{\ln(x+2)} =1 \ . $$

I'm in trouble. :/ Help please.

share|cite|improve this question
Thank You very much. Andrew, Hans Lundmark, Ragib Zaman. – user973511 Oct 24 '11 at 14:32
up vote 4 down vote accepted

Could you justify this: $$ \lim_{x\to +∞} \dfrac{\ln(x)}{\ln(x+2)} = \lim_{x\to +∞} \dfrac{\ln(x+2)-(\ln(x+2)-\ln(x))}{\ln(x+2)} = $$ $$ 1-\lim_{x\to +∞} \dfrac{\ln(1+\frac2x)}{\ln(x+2)} =1? $$

share|cite|improve this answer

Hint: Rewrite $\ln(x+2)$ as $\ln(x(1+2/x))=\ln x + \ln(1+2/x) = \left( 1 + \frac{\ln(1+2/x)}{\ln x}\right) \ln x$.

share|cite|improve this answer

Welcome to math.stackexchange! Does this satisfy you? $$ \frac{ \log x}{\log(x+2)} = \frac{\log x}{ \log x + \log(1+ 2/x) } = 1- \frac{\log (1+2/x)}{\log x + \log(1+2/x)} \to 1 .$$

share|cite|improve this answer

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.