Evaluate limit without L'Hopital's rule How can this limit be evaluated without the use of L'Hopital's rule. I already understand how to evaluate it with the use of it.

\begin{equation}
\lim_{x\to 0} (\frac{e^{3x}-1}{x})
\end{equation}

I am wondering if there is some sort of substitution I can use to prove it can evaluate to 3. Assume knowledge of first-principals.

This fact was used in the answer for this question.
Evaluating Limit Without L&#39;Hopital

I do not follow the solution to this question Evaluate $\lim_{x\to 0} \frac{a^x -1}{x}$ without applying L&#39;Hopital&#39;s Rule. If someone can expand further on that that would also be great!

I am not looking for similar answers to these: Show $\lim\limits_{h\to 0} \frac{(a^h-1)}{h}$ exists without l&#39;H&#244;pital or even referencing $e$ or natural log
Recall that by standard limit as $x\to 0$

$$\frac{e^x-1}{x}\to 1$$

then by $y=3x\to 0$

$$\frac{e^{3x}-1}{x}=3\frac{e^{3x}-1}{3x}=3\frac{e^y-1}{y}\to 3$$
Put $t=e^{3x} -1$ and find $x$ using log transformation.
You will be left with limit as $\frac{3t}{\ln(1+t)}$ then use log expansion you will get required answer.