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So I was looking through a Calculus book, in their first chapter section 2 about limits and I came across the following problem:

$$ \lim_{x\to 0} \dfrac{e^{3x}-e^x}{x}=2 $$

Immediately, I took the limit using L'Hopital's, but this is only the first chapter and they have yet to cover derivatives. Other than using Taylor Series (which would be taught later as well) is there another way to answer this question?

I had thought of some factorization

$$\dfrac{e^{3x}-e^x}{x}=\dfrac{e^x(e^{2x}-1)}{x}=\dfrac{e^x(e^x-1)(e^x+1)}{x}$$

Which made me think the problem could devolve into me showing that: $$ \lim_{x\to 0} \dfrac{e^x(e^x-1)}{x}=1 $$ and similarly, $$ \lim_{x\to 0} \dfrac{(e^x-1)}{x}=1. $$

However, even here I tend to think of Taylor series to prove this. Any help is immensely appreciated.

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    $\begingroup$ How do they define $e$? $\endgroup$ – mathworker21 Sep 1 '18 at 13:36
  • $\begingroup$ @farruhota: That was already noted, as well as the fact that $e^x\to 1.$ that's how the problem was reduced. $\endgroup$ – Cameron Buie Sep 1 '18 at 13:51
  • $\begingroup$ Have you met derivatives at this point and do you know what the derivative of $e^x$ is? (The limit you have reduced the problem to is the limit that gives the derivative of $e^x$ at $0$.) $\endgroup$ – Rob Arthan Sep 1 '18 at 13:57
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    $\begingroup$ Thank you, like I said I was trying to use only the content the book had used up to the point where the question was formulated. As I went back to it I found they defined $e$ as $\lim_{x\to 0} (1+x)^{1/x}$ which is had confused me. Seeing the limit $\lim_{t\to 0} \dfrac{e^t-1}{t}=1$ proof cleared up my answers. $\endgroup$ – katari Sep 1 '18 at 21:48
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Given the standard limit as $t \to 0$ $\frac{e^t-1}t \to 1$ we have that

$$\dfrac{e^{3x}-e^x}{x}=2e^x \frac{e^{2x}-1}{2x}\to2\cdot1\cdot1$$

To prove the standard limit recall that

$$e=\lim_{y\to \infty}\left(1+\frac1y\right)^y \implies 1=\log e=\lim_{y\to \infty}y\log \left(1+\frac1y\right) \implies \lim_{x\to 0} \frac{\log(1+x)}{x} \to 1 $$

therefore by $y=e^t-1 \to 0 \implies t=\log (1+y)$ we have

$$\frac{e^t-1}t=\frac{y}{\log (1+y)}\to 1$$

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  • $\begingroup$ What definitions and known facts is this answer based on? $\endgroup$ – Rob Arthan Sep 2 '18 at 0:02
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Your limit is equivalent to the well-known limit $\lim_{t\to 0}\frac{e^t-1}{t}=1$, but the most straightforward way to prove the last identity depends on your definition of the exponential function. For instance, if it is $$ e^x = \sum_{n\geq 0}\frac{x^n}{n!} $$ there is nothing to prove. If it is $e^x=\lim_{n\to +\infty}\left(1+\frac{x}{n}\right)^n$ you still have that in a neighbourhood of the origin $e^x=1+x+o(x)$, by Bernoulli's inequality or AM-GM.

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The question is how you have defined $e^x$. If it is as a power series, it is a bit silly to avoid Taylor series.

If you have defined the exponential as the inverse function of $\log x:=\int_1^x \frac1t dt$, then it is easy to show it is its own derivative and is $1$ at $0$

If you have defined $e^x=\lim_{n\to\infty}\left(1+\frac x n\right)^n$ you need to work slightly harder to show it is its own derivative and is $1$ at $0$, but it is not impossible.

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MVT:

1) x \ge 0.

$e^x -1 = {\displaystyle \int_{0}^{x}} e^t dt = e^{y}(x-0)$, where

$y \in [0,x]$.

$\lim_{x \rightarrow 0} \dfrac{e^x-1}{x} =$

$\lim_{x \rightarrow 0} e^{y(x)} = e^{0}$,

since the exponential function

is continuos and $\lim_{x \rightarrow 0} y(x)=0$.

2) $x \le 0$.

$1- e^x ={\displaystyle \int_{x}^{0}} e^{t} dt =$

$e^r (0-x)$, where $r \in [x,0].$

$\dfrac{e^x-1}{x}= e^r.$

Proceed to take the limit $x \rightarrow 0$, as in case 1).

3) Right limit = Left limit,

hence $\lim_{x \rightarrow 0} \dfrac{e^x-1}{x}=1$.

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  • $\begingroup$ Nice derivation Peter. The only observation is that the OP is requiring to do not use derivatives but that's of course a very nice and good answer! $\endgroup$ – user Sep 1 '18 at 21:51
  • $\begingroup$ Gimusi.Thanks, a reason for posting was that I have not seen the standard limit being derived by MVT for integrals. In the proof of this MVT, the one I looked at, differentiation is not used explicitly, rather integrability and IMV of continuos functions.Good to hear from you:) $\endgroup$ – Peter Szilas Sep 2 '18 at 6:32
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If $e$ is defined as $$e^x=\sum_{i=0}^{\infty}\frac{x^i}{i!}$$ Then the rest is immidiate, since $$\frac{e^x-1}{x}=\sum_{i=0}^{\infty}\frac{x^i}{(i+1)!}=1+\frac{x}{2}+\dots$$

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As you say f[x] = e^x[e^x-1][e^x+1]/x and

e^x[e^x -1]/x =1 as x tends towards 0

So your remaining factor [e^x +1] as x tends towards 0 is

e^0+1 = 2

f[x] =1[2] =2 as x tends towards 0

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  • $\begingroup$ It is correct but the big point here is to explain why $(e^x-1)/x \to 1$ holds whitout using derivatives. $\endgroup$ – user Sep 1 '18 at 21:53

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