real analysis - 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.

Answer

$$\lim_{x \rightarrow 0} \frac {ln(1+x)} x = \lim_{x \rightarrow 0} \frac 1 x ln(1+x) = \lim_{x \rightarrow 0} ln((1+x)^{\frac 1 x}) = ln(\lim_{x \rightarrow 0} (1+x)^{\frac 1 x}) = ln(e) = 1$$

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 $x=L$