Prove that $\lim_{x \to 0 } \frac{\ln(x+1)}{x} = 1$ I've looked around to see a proof for this limit and encountered this:
$$
\lim_{x \to 0 } \frac{\ln(x+1)}{x} 
$$
$$
\lim_{x \to 0 } \frac{1}{x} \ln(x+1)
$$
$$
\lim_{x \to 0 } \ln(x+1)^\frac{1}{x}
$$
$ t = \frac{1}{x}$
Then: 
$$
\lim_{t \to +- \infty } ln(\frac{1}{t}+1)^t
$$
$$
\lim_{t \to +- \infty } ln(e) = 1
$$
What I didn't understand is how did he transfer $\frac{1}{x} \ln(x+1)$ to this: $ \ln(x+1)^\frac{1}{x}
$
and how did he transfer this: $\ln(\frac{1}{t}+1)^t$ to this:
$
\ln(e) = 1
$
Is this the right approach to prove this limit? can someone explain me the steps with I didn't understand?
Thanks :)
 A: "What I didn't understand is how did he transfer $\frac{1}{x} \ln(x+1)$ to this: $ \ln(x+1)^\frac{1}{x}
$"
By the laws of logarithms, $\log a^b = b\log a$. This is just going the other way.
"and how did he transfer this: $\ln(\frac{1}{t}+1)^t$ to this:
$
\ln(e) = 1
$"
That's one definition of $e$, the base of the natural logarithm. $\displaystyle e = \lim_{t \to \infty} (1 + \frac{1}{t})^t$. Of course, $\ln e = 1$.
A: 
how did he transfer ${1\over x}ln(x+1)$ to this: $ln(x+1)^{1\over x}$

This is from a property of logarithm of exponents as can be seen here (look at "power" in the table).

how did he transfer this: $ln({1\over t}+1)^t$ to this: $ln(e)=1$

I would say they used $\lim_{t \to +- \infty } (\frac{1}{t}+1)^t = e)$.
  I'll admit it's been a little while for me so maybe someone else can verify the limit can "go through" the natural log.
My approach would have been to use L'Hopital's rule:  Simply take the derivative of the top and bottom separately and evaluate the limit.
Hope this helps.
A: $e^{a\ln(x)} = (e^{\ln(x)})^{a} = x^a = e^{\ln(x^a)}$, so definitely $a\ln(x) = \ln(x^a)$, because $e^x$ is injective function. 
$\lim_{t\rightarrow \infty} (1 + \frac{1}{t})^t = e$ by definition.
A: The reason $\frac{1}{x} \ln(x+1) = \ln(x+1)^\frac{1}{x}$  is true is because $a\ln(b)=\ln(b^a)$.
As for the second part, that's using the definiton of $e$ using limits, as shown here: http://mathworld.wolfram.com/e.html
An alternative way to find the limit is to use the Taylor series for $\ln(1+x)$ and dividing through by $x$, at which point you can just take the limit.
A: Since:
$$\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}$$
and $f(t)=\frac{1}{t}$ is a convex function, we have:
$$\frac{1}{n+\frac{1}{2}}\leq\log\left(1+\frac{1}{n}\right)\leq \frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right),\tag{1}$$
hence:
$$ \lim_{x\to 0}\frac{\log(1+x)}{x}=1$$
follows by squeezing. We have:
$$ a \log w = \log w^a$$
for any $w>0$.
A: It is perhaps worth pointing out that $\ln(x+1)^{1\over x}$ here means $\ln((x+1)^{1\over x})$, not $(\ln(x+1))^{1\over x}$.  As others have already observed, the equality with ${1\over x}\ln(x+1)$ comes from the general property for logarithms, $a\log b=\log(b^a)$.
As for the denouement, the general property is the following:  

If $\lim_{t\to c}g(t)=\ell$ and if $f$ is continuous at $\ell$, then
  $\lim_{t\to c}f(g(t))=f(\ell)$.

In this case, $c=\pm\infty$, $g(t)=({1\over t}+1)^t$, and $\ell$ turns out to be $e$ (for both $c=+\infty$ and $-\infty$).  The function $f$ is the natural logarithm, which is continous at $e$.  (Note, the subtlest point in the problem is that you have to take the limit in $t$ at both $+\infty$ and $-\infty$.)
A: Use L'Hospital's rule:
$$\lim_{x\to 0} \frac 1 x \ln (x+1) = \lim_{x \to 0} \frac {(\ln (x+1) )'}{(x)'}  = \lim_{x \to 0} \frac 1 {x+1} = 1$$
Another way:
Let $g:[1,x+1] \subset (0, +\infty) \to \mathbb R$ such that $g(u) = \ln u$ and $g'(u) = 1/u$. Then by MVT to $[1,x+1]$ it follows that $\exists k \in (1,x+1) $ so that $g'(k) = \frac 1 k = \frac {ln(x+1)}{x}$.
Finally $1 < k < x+1 \iff 1/x < 1/k < 1 \iff \frac 1 {x+1} <\frac {ln(x+1)} {x} < 1$. So, by squeezing you get that the limit is equal to 1. 
