find $\lim_{x \to \infty} \frac{\ln(x+1)}{x}$ I need to find: $$\lim_{x \to \infty} \frac{\ln(x+1)}{x}$$
without using L'Hospital, integrals or derivative!
I would like your help in order to know i'm not doing something invalid.
I was thinking of the following way:
$$
\lim_{x \to \infty } \frac{\ln(x+1)}{x}  =  \lim_{x \to \infty } \frac{1}{x} \ln(x+1)  =  \lim_{x \to \infty } \ln(x+1)^\frac{1}{x}
$$
and because $\frac{1}{x}$ tend to $0$ as $x$ approaches $\infty$, than $(x+1)^\frac{1}{x}$ tends to $1$ and $\ln(1) = 0$
and the limit is $0$.
But this way sounds off. I was trying also to express $x$ with $t$ and use $e$ (euler) somehow, but it didn't work.
Is my way valid? If not, do you have a better way without using L'Hospital's rule?
Thanks!

Edit:
While reading answers I was thinking of another way,
$$\lim_{x \to \infty} \frac{\ln(x+1)}{x}  =  \lim_{x \to \infty} \frac{\ln(x(1+\frac{1}{x}))}{x} = \lim_{x \to \infty} \frac{\ln(x)}{x} + \frac{\ln(1+\frac{1}{x})}{x} = $$ $$\lim_{x \to \infty} \ln(\sqrt[x]{x}) + \frac{\ln(1+\frac{1}{x})}{x}$$
now $\sqrt[x]{x}$ tends to $1$ as x approaches to $\infty$, hence $\ln(\sqrt[x]{x})$ tends to $0$ (is it valid to say so?)
and $\frac{\ln(1+\frac{1}{x})}{x}$ tends to $0$ is pretty obvious.
Am on missing something in this one?
Without L'Hospital Rule (or series for that matter!) you are left with using some basic properties of functions. For example set $\ln(x+1)=t$ so that $x=e^t-1$ and now you have an expression $\frac{t}{e^t-1}$ with $t $ going to infinity. To what extend does your teacher want you to prove that an exponential functions grows much faster than a linear term?
It is easy to show (See Here) that $\log(x)&lt;\sqrt{x}$ for all $x&gt;0$.  Therefore, we have
$$\frac{\log(1+x)}{x}\le \frac{\sqrt{x+1}}x$$
And you can finish now.