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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

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    $\begingroup$ It's literally the definition of a derivative at $0$: $f'(0)$, for $f(x) = e^{3x}$. $\endgroup$
    – Clement C.
    Commented Jun 24, 2018 at 6:08
  • $\begingroup$ use Taylor formula for $f(x)=\exp^{3x}-1$ till second order $\endgroup$
    – markovchain
    Commented Jun 24, 2018 at 6:08

4 Answers 4

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As @Clement C. points out, $$\lim_{x\to0}\frac {e^{3x}-1}x=\lim_{x\to0}\frac {e^{3x}-e^{3\cdot 0}}{x-0}=(e^{3x}){^{'}}(0)=(3e^{3x})(0)=3e^{3\cdot 0}=3e^0=3\cdot 1=3$$.

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Recall that by standard limit as $x\to 0$

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

then by $y=3x\to 0$

$$\frac{e^{3x}-1}{x}=3\frac{e^{3x}-1}{3x}=3\frac{e^y-1}{y}\to 3$$

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  • $\begingroup$ One of the answers I've linked uses this property. I am asking why $\endgroup$
    – Uday
    Commented Jun 24, 2018 at 6:12
  • $\begingroup$ @Uday It is a simple agebraic manipulation to obtain a limit in the standard form and then we can use the well known result, indeed $3x\to 0$ $\endgroup$
    – user
    Commented Jun 24, 2018 at 6:14
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One way to evaluate the limit is using Taylor series:

The Taylor series at $x=0$ of $\frac{e^{3x}-1}{x}$ is $$ 3 + \frac{9x}{2} + \frac{9x^2}{2} + \frac{27x^3}{8} + \frac{81x^4}{40} + O(x^5) $$

So $$\lim_{x\to 0} \left(\frac{e^{3x}-1}{x}\right) = 3$$

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    $\begingroup$ It is a derivative. Why do you do an expansion to order $x^4$?! $\endgroup$
    – Clement C.
    Commented Jun 24, 2018 at 6:37
  • $\begingroup$ @ClementC. Yeah, but it's another way to evaluate the limit if you don't realize it's a derivative at first glance. $\endgroup$
    – F.A.
    Commented Jun 24, 2018 at 7:02
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    $\begingroup$ But then, a Taylor expansion to order $x$ suffices. $\endgroup$
    – Clement C.
    Commented Jun 24, 2018 at 15:26
  • $\begingroup$ @ClementC.Sure, but it was easy to calculate and I was feeling generous. $\endgroup$
    – F.A.
    Commented Jun 24, 2018 at 20:39
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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.

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