# How can I determine this limit's value?

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.

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Thank You very much. Andrew, Hans Lundmark, Ragib Zaman. –  user973511 Oct 24 '11 at 14:32

Could you justify this: $$\lim_{x\to +∞} \dfrac{\ln(x)}{\ln(x+2)} = \lim_{x\to +∞} \dfrac{\ln(x+2)-(\ln(x+2)-\ln(x))}{\ln(x+2)} =$$ $$1-\lim_{x\to +∞} \dfrac{\ln(1+\frac2x)}{\ln(x+2)} =1?$$
Hint: Rewrite $\ln(x+2)$ as $\ln(x(1+2/x))=\ln x + \ln(1+2/x) = \left( 1 + \frac{\ln(1+2/x)}{\ln x}\right) \ln x$.
Welcome to math.stackexchange! Does this satisfy you? $$\frac{ \log x}{\log(x+2)} = \frac{\log x}{ \log x + \log(1+ 2/x) } = 1- \frac{\log (1+2/x)}{\log x + \log(1+2/x)} \to 1 .$$