Limit of $\dfrac{t}{\ln(1+t)}$ without L'Hospital I'm trying to prove: $\lim_{t\rightarrow 0} \dfrac{t}{\ln(1+t)}=1$ without use L'Hospital and derivate, because I still do not define derived from the logarithm, is there any way to prove it by the definition epsilon-delta? So far I have only defined $\log$ as the inverse function of $a^{x}$ and particular case $a = e$ although the latter is proposed very weak, it may be improved later using integrals, some help thanks in advance.
 A: $$\lim_{x \rightarrow 0} \frac {ln(1+x)} x = \lim_{x \rightarrow 0} \frac 1 x ln(1+x) = \lim_{x \rightarrow 0} ln((1+x)^{\frac 1 x}) = ln(\lim_{x \rightarrow 0} (1+x)^{\frac 1 x}) = ln(e) = 1$$
We use the fact that $a ln(b) = ln(b^a)$
Edit: Based on comments, I added one last step before "ln(e)". This is to showcase that limits have the property that $\lim_{x \rightarrow x_0} f(g(x)) = f(\lim_{x \rightarrow x_0} g(x) )$ provided that $\lim_{x \rightarrow x_0} g(x) = L, L \in \mathbb R$ and that $f$ is continuous at $x=L$
A: Of course we need a definition of $\ln$, otherwise we can prove nothing about it.  Perhaps the definition is
$$
\ln x = \int_1^x\frac{ds}{s}
$$
Case 1: $t \to 0^+$.
$$
\frac{\ln(1+t)}{t} = \frac{1}{t}\int_1^{1+t}\frac{ds}{s}
$$
Now when $1 \le s \le 1+t$, we have $\frac{1}{1+t}\le \frac{1}{s}\le 1$.
Therefore
$$
\frac{1}{t}\int_1^{1+t}\frac{ds}{s} \le \frac{1}{t}\int_1^{1+t} ds = \frac{1}{t}\;t = 1
$$
and
$$
\frac{1}{t}\int_1^{1+t}\frac{ds}{s} \ge \frac{1}{t}\int_1^{1+t}\frac{ds}{1+t}
=
\frac{1}{t}\;t\;\frac{1}{1+t} = \frac{1}{1+t}.
$$
Thus
$$
\frac{1}{1+t} \le \frac{\ln(1+t)}{t} \le 1
$$
and take limit to get
$$
1 \le \lim_{t\to 0}\frac{\ln(1+t)}{t} \le 1 .
$$
Case 2: $t \to 0^-$ is similar.
A: If you grant the definition of $$e=\lim_{t\rightarrow 0}(1+t)^{1/t},$$ the problem is equivalent to showing $$\lim_{t\rightarrow 0}\frac {\ln(1+t)}t$$ $$=\lim_{t\rightarrow 0}\ln(1+t)^{\frac 1 t}=1,$$ by the definition of $e$ and by the continuity of $\ln$.
A: Along the same lines as GEdgar's answer (which was posted while I was working on this), we have
$$1-{\ln(1+t)\over t}={1\over t}\int_0^t\left(1-{1\over1+x}\right)dx={1\over t}\int_0^t{x\over1+x}dx=t\int_0^1{u\over1+tu}du$$
so, for $t\gt-1$ (so that $1+tu\gt0$ for all $u\in[0,1]$),
$$\left|1-{\ln(1+t)\over t}\right|=|t|\int_0^1{u\over1+tu}du\to0$$
since
$$\int_0^1{u\over1+tu}du\to\int_0^1u\,du={1\over2}$$
or, if you prefer,
$$0\lt\int_0^1{u\over1+tu}du\lt\int_0^1{1\over1-|t|}du={1\over1-|t|}$$
(assuming $-1\lt t\lt1$), so that
$$0\le|t|\int_0^1{u\over1+tu}du\lt{|t|\over1-|t|}\to0$$
A: Hint: Use the power series for $ \ln$
\begin{eqnarray*}
\ln(1+t) = \sum_{n=1}^{\infty} (-1)^{n+1}  \frac{t^n}{n}.
\end{eqnarray*}
A: $$\lim\limits_{t\to0}\frac{\ln(1+t)}t=\lim\limits_{t\to0}\frac{\ln(1+t)-\ln1}t$$
That's the definition of the derivative of $\frac d{dx}\ln x$ evaluated at $x=1$, which is obviously $\frac11=1$.  Therefore, 
$$\lim\limits_{t\to0}\frac t{\ln(1+t)}=1$$
