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I have problem with $$\lim_{x\to 0}\frac {e^{3x}-1}{e^{x}-1} $$ I have no idea what to do first.

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    $\begingroup$ L'Hopitals rule.. See here for a similar problem. $\endgroup$ Commented Jan 6, 2016 at 18:00
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    $\begingroup$ @Mattos There's a much more fundamental solution in the Answers section. $\endgroup$ Commented Jan 6, 2016 at 18:06
  • $\begingroup$ @FlybyNight And? $\endgroup$ Commented Jan 6, 2016 at 18:07
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    $\begingroup$ @Mattos : L'Hopital's rule is very efficient but often affords little or no insight. ${}\qquad{}$ $\endgroup$ Commented Jan 6, 2016 at 18:10
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    $\begingroup$ . . . . . and "solve" is the wrong word. One solves problems; one solves equations; one evaluates expressions. "Find" or "evaluate" would be appropriate. There are a number of words in mathematics that, in the hands of the inexperienced, get used whenever they don't know which word to use, and "solve" is one of those. ${}\qquad{}$ $\endgroup$ Commented Jan 6, 2016 at 18:14

7 Answers 7

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without using L'Hopital's rule

$$\lim_{x\to 0}\frac{e^{3x}-1}{e^x-1}=\lim_{x\to 0}\frac{(e^x-1)(e^{2x}+e^x+1)}{e^x-1}=\lim_{x\to 0} e^{2x}+e^x+1=3$$

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  • $\begingroup$ I changed $lim{x\to0}$ to $\displaystyle\lim_{x\to0}$. ${}\qquad{}$ $\endgroup$ Commented Jan 6, 2016 at 18:29
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Hint Recognize $e^{3x}-1$ as a difference of cubes.

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  • $\begingroup$ Very nice answer. $\endgroup$ Commented Jan 6, 2016 at 18:06
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HINT: Note that $x^3-1=(x-1)(x^2+x+1)$

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    $\begingroup$ Great answer, and no need for L'Hopital's Rule. Perhaps use a different variable than the one already in use in the original question? For example $y^3-1 \equiv (y-1)(y^2+y+1)$. $\endgroup$ Commented Jan 6, 2016 at 18:03
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Hint: Use equivalents: $$\mathrm e^{ax}-1\sim_0 ax,\quad\text{hence}\quad \frac{\mathrm e^{ax}-1}{\mathrm e^x-1}\sim_0 \frac{ax}x=a.$$

Alternative hint: $$\frac{\mathrm e^{ax}-1}{x}\xrightarrow[x\to0]{}(\mathrm e^{ax})'\,\Big\lvert_{x=0}$$

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    $\begingroup$ This answer needs more explanation. I know about limits, but don't really understand what you're doing here. What does $\sim_0$ mean? How is the limit of the fraction $$\frac{\mathrm{e}^{ax}-1}{x}$$ related to the derivative of $\mathrm{e}^{ax}$? $\endgroup$ Commented Jan 6, 2016 at 18:08
  • $\begingroup$ @Fly by Night: The problem is with the first hint? $\endgroup$
    – Bernard
    Commented Jan 6, 2016 at 18:09
  • $\begingroup$ This answer does not need to be improve, everything is crystal-clear. Besides, the OP has shown no effort. @FlybyNight there is no need to be aggresive. $\endgroup$
    – C. Falcon
    Commented Jan 6, 2016 at 18:14
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    $\begingroup$ @Fly by Night: I've added a link (different from Clement C's, more specifically about equivalence) hopefully useful about this important notion in analysis. $\endgroup$
    – Bernard
    Commented Jan 6, 2016 at 18:28
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    $\begingroup$ @Bernard That's great, thank you! $\endgroup$ Commented Jan 6, 2016 at 18:34
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Let's apply L'Hospital's Rule here; i.e.,

$$\lim_{x\to 0}\frac {e^{3x}-1}{e^{x}-1}= \lim_{x\to 0}\frac {3e^{3x}}{e^{x}}=\bbox[5px,border:2px solid #F0A]3\,.$$

OR We can try substitution if you're not familiar with L'H rule:

$$t=e^x\Rightarrow \color{blue}{t\rightarrow1 \: as \: x\rightarrow0} \: \Rightarrow\lim_{x\to 0}\frac {e^{3x}-1}{e^{x}-1}=\lim_{t\to 1}\frac {t^3-1}{t-1}=\lim_{t\to 1}\frac {(t-1)(t^2+t+1)}{t-1}=\lim_{t\to 1}(t^2+t+1)=\bbox[5px,border:2px solid #F0A]3$$

What you see above is the graph of $y=\frac{t^3-1}{t-1}$. As you can see it is not defined at $t=1$, but it has a limit that is equal to $3$ at $t=1$.

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    $\begingroup$ L'Hopital's rule is often very efficient but often gives little or no insight. $\endgroup$ Commented Jan 6, 2016 at 18:15
  • $\begingroup$ I agree, that's why I provided OP with another way to solution. $\endgroup$
    – frosh
    Commented Jan 6, 2016 at 18:16
  • $\begingroup$ Why do you put the sentence-ending period at the beginning of the next line after the sentence ends? I fixed that in an edit, but then you changed it back. $\endgroup$ Commented Jan 6, 2016 at 18:20
  • $\begingroup$ I did not realize. You're free to edit again if it bothers you. $\endgroup$
    – frosh
    Commented Jan 6, 2016 at 18:21
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My first thought was to write it as $$ \frac{\left(\lim\limits_{x\to 0} \dfrac{e^{3x}-e^{3\,\cdot\,0}}{x-0} \right)}{\left(\lim\limits_{x\to 0} \dfrac{e^x-e^0}{x-0} \right)} = \frac{\left.\dfrac d {dx} e^{3x} \,\right|_{x=0}}{\left.\dfrac d {dx} e^x \,\right|_{x=0}} = \text{etc.} $$

But the simpler way is to write $e^{3x}-1 = (e^x-1)(e^{2x}+e^x+1)$ and then cancel.

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Looking at Taylor series for $x \to 0$: $$ \lim_{x \to 0} \frac{e^{3x}-1}{e^x-1} = \frac{1 + 3x - 1 + \mathcal{O}(x^2)}{1 + x - 1 + \mathcal{O}(x^2)} = \frac{3x + \mathcal{O}(x^2)}{x + \mathcal{O}(x^2)} = 3 $$

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