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?
 A: 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?
A: Let $c\in\mathbb{R}$ and $f(x)=\log(x)-\sqrt{x}+c$, then $f'(x)=\frac{1}{x}-\frac{1}{2\sqrt{x}}=\frac{2-\sqrt{x}}{2x}$, therefore the maximum of $f$ is $f(4)=\log 4-2+c$. We take $c=2-\log 4$, we thus have $f(x)\leqslant 0$ for all $x>0$, which leads to $\frac{\log(x)}{x}\leqslant\frac{1}{\sqrt{x}}+\frac{\log 4-2}{x}\underset{x\rightarrow +\infty}{\longrightarrow}0$.
A: I will use the substitution $y=\ln(1+x)$.
Let $n$ be $n=[y]$
Then $x=e^y-1$. We can simply estimate for $x>10$ (say):
$$
0\le \frac{\ln(x+1)}x=\frac y{e^y-1}<\frac y{e^y/2}<\frac y{2^y/2}
<
\frac {n+1}{2^{n-1}}
=
\frac {n+1}{(1+1)^{n-1}}
<
\frac {n+1}{\binom{n-1}2}
<\frac {2n}{\frac n2\cdot\frac n2\cdot \frac 12}
=\frac {16}n\to 0
\ .
$$
A: It is easy to show (See Here) that $\log(x)<\sqrt{x}$ for all $x>0$.  Therefore, we have
$$\frac{\log(1+x)}{x}\le \frac{\sqrt{x+1}}x$$
And you can finish now.
A: $\ln\nearrow\, $ thus for $\, x\gg 1\, $ then $\, x+1<2x\implies 0<\dfrac{\ln(x+1)}{x}<\dfrac{\ln(2x)}{x}=\underbrace{\dfrac{\ln(2)}{x}}_{\to 0}+\dfrac{\ln(x)}{x}$
So we have reduced the problem to studying $f(x)=\dfrac{\ln(x)}{x}$
$f'(x)=\dfrac{1-\ln(x)}{x^2}<0$ at infinity, so $f\searrow$ at infinity and in particular this means that $f$ is bounded by some $M$ at infinity.
Now we can use the multiplicative to additive property of logarithm,
$f(x)=\dfrac{2\ln(\sqrt{x})}{(\sqrt{x})^2}<\dfrac{2M}{\sqrt{x}}\to 0$
