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.

$$\lim\limits_{t\to0}\frac{\ln(1+t)}t=\lim\limits_{t\to0}\frac{\ln(1+t)-\ln1}t$$
That's the definition of the derivative of $\frac d{dx}\ln x$ evaluated at $x=1$, which is obviously $\frac11=1$.  Therefore, 
$$\lim\limits_{t\to0}\frac t{\ln(1+t)}=1$$
