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I have to solve this limit:

$$\lim\limits_{x\to 0} \left( \dfrac{e^x - 1 }{x} \right) $$

I can easily do it by using hospital's rule, numerator becomes $e^0$ and denominator becomes $1$.

How should I solve that limit by using definition of $e$ as $e= \lim\limits_{n \to \infty } {\left( {1 + \frac 1n } \right)}^n$

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  • $\begingroup$ Let n= 1/x. Then "$x\to 0$" becomes "$n\to\infty$". $\endgroup$ – user247327 Mar 7 '17 at 14:58
  • $\begingroup$ Can you use the series expansion of $e^{x}$? $\endgroup$ – mrnovice Mar 7 '17 at 14:58
  • $\begingroup$ You would use L'Hopital (involving computing the derivative of $\exp$ at $0$, and invoking a rather non-trivial theorem to conclude) to compute what is by definition the derivative of $\exp$ at $0$? $\endgroup$ – Clement C. Mar 7 '17 at 15:01
  • $\begingroup$ See this one $\endgroup$ – Juniven Mar 7 '17 at 15:16
  • $\begingroup$ @ΘΣΦGenSan yes,I wanted to solve by using definition of x given in my question $\endgroup$ – Noddy Mar 7 '17 at 15:20
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Method 1: Defintion of derivative Hint:

$$\lim_{x\to 0}\frac{e^x-e^0}{x-0}.$$

This is nothing but $\frac{d}{dx}e^x|_{x=0}$.


Method 2:Taylor Series

You could also plug in the definition of $e^x$ as a Taylor series and collect all higher order terms as $O(x^2)$: $e^x=1+x+O(x^2)$.


Method 3: Euler

The last method was the method Euler used, which is not rigorous but still very nice.

Look at $$\lim_{x\to 0}\frac{a^x-1}{x}$$, Euler first showed convergence (not sure if he even cared about it :D) and then said that we want to choose $a$ such that the limit is $1$. So approximately $$\frac{a^x-1}{x}\approx 1\implies a^x\approx 1+x \implies a \approx (1+x)^{1/x}.$$ Now, substitute $n=1/x$, hence $n\to \infty$ and we get $$a\approx \lim_{n\to \infty}\left(1+\frac{1}{n} \right)^n=e$$

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You can see that:

$$\frac{e^x-1}{x}=\frac{e^x-e^0}{x-0}.$$

So since $x\mapsto e^x$ is differentiable,

$$\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{x\to 0}\frac{e^x-e^0}{x-0}=\lim_{x\to 0}\frac{d}{dx} (e^x)=\lim_{x\to 0}e^x=1.$$

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    $\begingroup$ This is basically a proof of l'Hospital rule. It does not answer OP's question... $\endgroup$ – Roberto Rastapopoulos Mar 7 '17 at 15:01
  • $\begingroup$ Also, this is basically wrong. You are using continuity of the derivative of $\exp$, for which reason? (1) you do not need it, as this is the definition of $\exp'(0)$, (2) the same argument ought to work for functions whose derivative at $0$ exists, but who are not continuously differentiable. $\endgroup$ – Clement C. Mar 7 '17 at 15:04
  • $\begingroup$ No. This shows the limit it f'(x) while LH says that f/g has the same limit as f'/g'. $\endgroup$ – marty cohen Mar 7 '17 at 15:06
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    $\begingroup$ @martycohen Roberto's comment is wrong indeed, but so is the step which writes $\lim_{x\to 0} \frac{d}{dx} e^x$ (it is only true here because $\exp$ is nice, but replacing $\exp$ by some $f$ differentiable at $0$ (and nowhere else) should still be OK in the OP's question). It should be $\frac{d}{dx} e^x \mid_{x=0}$, or even better $\exp'(0)$. $\endgroup$ – Clement C. Mar 7 '17 at 15:09
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Here is an attempt to use only integers.

To show $\lim_{x \to 0} \dfrac{e^x-1}{x} = 0 $ using $e = \lim_{n \to \infty} \left(1+\frac1{n}\right)^n $.

From this, $e^x = \lim_{n \to \infty} \left(1+\frac1{n}\right)^{nx} $.

This is a double limit, which is often problematical.

For general $x$, this seems to require the generalized binomial theorem. To avoid this, I will assume $x = \frac1{m}$ as integer $m \to \infty$, and let $n=km$ as integer $k \to \infty$.

Then $e^{1/m} = \lim_{k \to \infty} \left(1+\frac1{mk}\right)^{k} $. Now we can use the binomial theorem.

$\begin{array}\\ \left(1+\frac1{mk}\right)^{k} &=\sum_{j=0}^k \binom{k}{j}\frac1{(mk)^j}\\ &=\sum_{j=0}^k \frac{\prod_{i=0}^{j-1}(k-i)}{j!}\frac1{(mk)^j}\\ \text{so}\\ f(m, k) &=\dfrac{\left(1+\frac1{mk}\right)^{k}-1}{1/m}\\ &=m\sum_{j=1}^k \frac{\prod_{i=0}^{j-1}(k-i)}{j!}\frac1{(mk)^j}\\ &\le m\sum_{j=1}^k \frac{1}{j!}\frac1{m^j}\\ &= \sum_{j=1}^k \frac{1}{j!}\frac1{m^{j-1}}\\ &=1+ \sum_{j=2}^k \frac{1}{j!}\frac1{m^{j-1}}\\ &\le 1+ \frac1{m}\sum_{j=2}^k \frac{1}{j!}\frac1{m^{j-2}}\\ &= 1+ \frac1{m}\sum_{j=0}^{k-2} \frac{1}{(j+2)!}\frac1{m^j}\\ \end{array} $

Since $\sum_{j=0}^{k-2} \frac{1}{(j+2)!}\frac1{m^j}$ is easily shown to be bounded by $1$ (actually, by $\sum_{j=0}^{k-2} \frac{1}{(j+2)!} = e-2$), $f(m, k) \le 1+\frac1{m}$.

Since $f(m, k) \ge 1$ (from its first term), $\lim_{m \to \infty} f(m, k) = 1$.

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