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I have a homework problem which asks me to show $$ \lim\limits_{x \to 0}\frac{e^x-1}{x} = 1 $$

Any help is appreciated.

Jeremy Upsal
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Diminshka1995
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4 Answers4

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By the definition of the derivative and if we denote $f(x)=e^x$ then $$\lim_{x\to0}\frac{e^x-1}{x}=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=f'(0)=e^0=1$$

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We have $e^x=1+x+\frac{x^2}{2}+O(x^3)$, so $$\frac{e^x-1}{x}=\frac{x+x^2/2+O(x^3)}{x}=1+x/2+O(x^2)$$

Taking limits as $x\to 0$, we get $1+0=1$.

vadim123
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By L'Hôpital's rule $$\lim\limits_{x \to 0}\frac{e^x-1}{x} = \lim\limits_{x\to0}\frac{e^x}{1} = 1$$

user85798
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    Don't you already need to know the derivative of $e^x$ to use L'Hopital here? – GEdgar Nov 12 '13 at 19:51
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    Yes.. pretty simple derivative – user85798 Nov 12 '13 at 20:06
  • @OliverBel: but the limit you're supposed to calculate is precisely the value of the derivative at zero... This is the same mistake as using the rule to calculate the limit of $\sin(x)/x$ at $0$. – tomasz Nov 14 '13 at 21:02
  • So? What's wrong with assuming we know the derivative? We choose $e$ to have that property. – user85798 Nov 15 '13 at 02:26
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You can use l'Hopitals rule. Notice that if we substitute directly we get $\frac{0}{0}$, an indeterminate form. So we can differentiate the numerator and denominator separately then take the limit. Then we have

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

Though I've doubt you've seen this yet, we don't even need that because we can use the Taylor series for $e^x$.

$$\lim_{x \to 0}\frac{e^x-1}{x}=\lim_{x\to 0} \frac{\sum_{n=0}^\infty\frac{x^n}{n!}-1 }{x}=\lim_{x \to 0} \frac{1+\sum_{n=1}^\infty \frac{x^n}{n}-1}{x}=\lim_{x \to 0}\frac{\sum_{n=1}^\infty \frac{x^n}{n!}}{x}=\lim_{x\to 0} \sum_{n=1}^\infty \frac{x^{n-1}}{n!}$$

But then we have

$$\lim_{x\to 0} \sum_{n=1}^\infty \frac{x^{n-1}}{n!}=\sum_{n=1}^\infty \frac{0^{n-1}}{n!}=1$$

Though seeing that last step with $0^0=1$ is a bit trickier. Not necessarily the way to do it but another way of thinking of it for those that know the Taylor Series.

EDIT: You wouldn't even need to use the fact that $0^0$ is $1$ as

$$ \lim_{x\to 0} \sum_{n=1}^\infty \frac{x^{n-1}}{n!}=\lim_{x\to 0} 1+\sum_{n=2}^\infty \frac{x^{n-1}}{n!}=1+ \sum_{n=2}^\infty \frac{0^{n-1}}{n!}=1 $$

mathematics2x2life
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  • I'm confused to as why my or vadim123's solutions using Taylor Series are so despised, especially since I also did it by l'Hopitals. If the down voters could elaborate so I can improve answers in the future that would be most helpful! – mathematics2x2life Nov 12 '13 at 21:37
  • Haters gonna hate. Some people define $e^x$ with Taylor series; then our solutions are spot on. – vadim123 Nov 13 '13 at 15:31
  • @vadim123: Your solution is correct under this assumption (so I'm not downvoting it, unlike this and Oliver Bel's), using L'Hospital rule is circular no matter how you define the function. Of course, it's OP's fault for not singling out a definition of $e^x$ he's using (for instance, if you define it as the solution of $y'=y$ with $y(0)=1$, this is trivial, and it makes Sami's answer sort of valid), but I can't think of any way to justify using L'Hospital rule for this. Can you? – tomasz Nov 14 '13 at 21:08
  • @tomasz I fail to see your logic here. vadim123's solution is no different than my alternate solution to l'Hopitals other than he truncates the series after a few terms whereas I continue to work with it in its entirety. Moreover, I give the solution using l'Hopitals rule. If you don't allow solutions using l'Hopitals rule or Taylor series or using the idea that you may be taking the derivative of a certain function, I fail to see what kind of solution you wanted. It makes no sense to find fault with every possible approach saying they aren't rigorous without suggesting a method that might be. – mathematics2x2life Nov 14 '13 at 21:16
  • @mathematics2x2life: it's hard to tell what a proper answer for this question would be, as OP did not define $e^x$, as I have said, but using L'Hospital is circular, as you are using the entire derivative to calculate the value of the derivative at zero, which is circular. Whether or not a solution using the "Taylor expansion" is correct would depend on whether you take it as the definition or the exponential function or as the Taylor expansion of $e^x$ defined otherwise. Circular logic is incorrect no matter how you look at it, though. – tomasz Nov 15 '13 at 01:09
  • @mathematics2x2life: as for the correct solution, depending on your definition of $e^x$, vadim's solution may be correct, or Sami's might be, or one of the ones in the question this one is marked a duplicate of. – tomasz Nov 15 '13 at 01:13
  • @tomasz To require a definition of $e^x$ to do a problem as simple as this seems overly pedantic. I highly doubt when you first learned l'Hopitals rule that questions had to be phrased as you are desribing. Moreover, you can view this as calculating a derivative or you can view it as a simple limit problem. The OP may not have specified much, but had he/she not been able to use the derivative concept, I'm sure the question would have been phrased in terms of $\epsilon-\delta$'s instead. – mathematics2x2life Nov 15 '13 at 01:14
  • @tomasz Moreover, it doesn't matter how you define $e^x$ as you can show the possible characterizations of $e^x$ result in the same function. Choosing one over the other depends on its utility in the problem and preference. You can't say one isn't correct and have another be--they are equivalent! – mathematics2x2life Nov 15 '13 at 01:17
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    @mathematics2x2life: but the limit is, by definition, the value of derivative! If someone asked how to calculate the volume of a $4$-ball of radius $R$, would you consider an answer that says that $\pi^2R^4/2$ because the volume of an $n$-ball is $\frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)}R^n$ acceptable? I wouldn't, similarly I don't think your answer is acceptable. They are equivalent, but it takes some work to show that they are equivalent -- including calculating the derivative. – tomasz Nov 15 '13 at 01:19
  • Work or not they are equivalent! So no matter what definition you use, the answers remain the same as they are equivalent! Moreover, to demand that you require such a high level of mathematical knowledge to do so basic a problem, you ask too much of the student! You wouldn't tell a high schooler that his calculation $\int e^x dx= e^x+C$ is incorrect because he hasn't given his definition of $e^x$ and also because he didn't prove it using $\epsilon-\delta$'s. – mathematics2x2life Nov 15 '13 at 01:26
  • I'm sad that the question is closed. I would to have loved to see your solution to see what you are asking for since you've only said what you didn't want to see versus what you think is appropriate. – mathematics2x2life Nov 15 '13 at 01:27