# Find $\lim_{x\to \infty} \ln(x+1)/(\ln(x))$

I have to solve: $\lim\limits_{x\to \infty} \frac{\ln(x+1)}{\ln(x)}$. Can you give me any hints to go? Thanks a lot!

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$$\lim\limits_{x\to +\infty} \frac{\ln\left(x+1 \right)}{\ln(x)}=\lim\limits_{x\to +\infty} \frac{\ln\left(x\left(1+\frac{1}{x}\right)\right)}{\ln(x)}=\lim\limits_{x\to +\infty} \frac{\ln x+\ln \left(1+\frac{1}{x}\right)}{\ln(x)}=1$$

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L’Hospital’s rule is a perfectly good, straightforward way to evaluate the limit, and in this case it’s easy; there’s no reason not to use it. However, there is also a pretty simple way to get it more directly. Start by rewriting the numerator:

$$\ln(x+1)=\ln x\left(1+\frac1x\right)=\ln x+\ln\left(1+\frac1x\right)\;.$$

I’ll spoiler-protect the rest to give you a chance to work it out if you want; mouse-over to see it.

$\displaystyle\frac{\ln(x+1)}{\ln x}=\frac{\ln x+\ln\left(1+\frac1x\right)}{\ln x}=1+\frac{\ln\left(1+\frac1x\right)}{\ln x}$, whose limit as $x\to\infty$ is easy to evaluate.

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hint:use L'Hôpital's rule to solve it

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$$\lim_{x\to \infty} \ln(x+1)/\ln(x)=\lim_{x\to \infty} \ln(x)/\ln(x)=1$$