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'Hopital
I do not follow the solution to this question Evaluate $\lim_{x\to 0} \frac{a^x -1}{x}$ without applying L'Hopital'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'Hôpital or even referencing $e$ or natural log
Best Answer
As @Clement C. points out, $$\lim_{x\to0}\frac {e^{3x}-1}x=\lim_{x\to0}\frac {e^{3x}-e^{3\cdot 0}}{x-0}=(e^{3x}){^{'}}(0)=(3e^{3x})(0)=3e^{3\cdot 0}=3e^0=3\cdot 1=3$$.