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I can only get $$\lim_{x \to 0} \frac{e^x - 1}{x}$$ to be undefined since $\frac{0}{0}$ doesn't mean anything. How can I manipulate the expression to get a valid answer?

Bérénice
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user348264
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  • exp(x) = 1+x+0.5x^2+..... – Cardinal Jun 16 '16 at 20:15
  • I suggest you to take a look here: https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule – Onil90 Jun 16 '16 at 20:16
  • What do you know about the exponential function? – Git Gud Jun 16 '16 at 20:16
  • Such a basic limit depends essentially on how you define the exponential function. – egreg Jun 16 '16 at 20:17
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    @Dman Using L'H to compute a limit $(f(x)-f(0))/x$ when $x\to0$ is making a logical circle since the ratio converges to $f'(0)$ by definition. – Did Jun 16 '16 at 20:20
  • @Onil90 No thanks, see previous comment. – Did Jun 16 '16 at 20:20
  • @Did yes you are right, thanks. – fosho Jun 16 '16 at 20:21
  • There are many questions about the evaluation of this limit on MSE and many times OP does not include a definition of $e^{x}$ in the question. Answer to the question essentially depends on how you define $e^{x}$ because there are several accepted approaches to define $e^{x}$ and develop a theory of exponential (and logarithmic functions). Please include a definition of $e^{x}$ in your question. See one approach to this question at http://math.stackexchange.com/a/541330/72031 – Paramanand Singh Jun 17 '16 at 07:18
  • @Did very nice observation!! – Onil90 Jun 17 '16 at 07:30

6 Answers6

5

Use the Taylor series of the exponential :

$$e^x=\sum_0^{\infty}\frac{x^n}{n!}$$

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

You can also recognise the definition of the derivative of a function $f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x}$, here $f(x)=e^x$, so :

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

Bérénice
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This is nothing more than

$$\lim_{x\to 0} \frac{f(x) - f(0)}{x-0}$$

for $f(x) = e^x.$ Look familiar?

zhw.
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  • Sometimes this limit needs to be evaluated in this very context. How do you do it then? (The answer depends on your definition of $e^x$, of course; also, this question is mostly rhetorical) – MT_ Jun 16 '16 at 20:27
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    You do whatever it takes to evaluate it. Under no circumstances would you ever use the result you are trying to prove in the proof itself. – zhw. Jun 16 '16 at 20:29
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Without L'Hospital's rule we can write as follows: Since $ \lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e$

$${ e }^{ x }-1=t\Rightarrow x=\ln { \left( t+1 \right) } \\ \lim _{ t\rightarrow 0 }{ \frac { t }{ \ln { \left( t+1 \right) } } } =\lim _{ t\rightarrow 0 }{ \frac { 1 }{ \ln { { \left( t+1 \right) }^{ \frac { 1 }{ t } } } } } =\frac { 1 }{ \ln { e } } =1\\ $$

haqnatural
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You can do a series expansion, or you can use l'Hopital's rule. But, If you don't know those yet.

$e^x = \lim_\limits{n\to\infty} (1+\frac xn)^n$

So, you can do a binomial expansion on that for a finite $n.$ Find your limit. And then look at the implications as $n$ goes to infinity.

Doug M
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  • I like this answer -- the reasonable cases where this limit is evaluated is when you want to know the derivative of $e^x$. Using material that assumes the knowledge of this result to prove it is often circular under this common context. – MT_ Jun 16 '16 at 20:29
  • This is a good approach but is tricky and not as simple as one might think. See http://math.stackexchange.com/a/541330/72031 – Paramanand Singh Jun 17 '16 at 07:20
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From the definition of $e$, we can easily obtain the following inequality

\begin{equation} (1+x)^{\large{1\over x}}\le e\le(1-x)^{-\large{1\over x}} \end{equation}

then applying the sandwich theorem, we obtain

\begin{equation} 1\le \lim_{x\to0}\ {e^x-1\over x}\le \lim_{x\to0}\ {1\over 1-x}=1 \end{equation}

Sophie Agnesi
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You can use L'Hôpital's rule

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

Oriol
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    Really it's a bit ridiculous to do that. The limit is nothing more than the definition of the derivative of $e^x$ at $0.$ – zhw. Jun 16 '16 at 20:17
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    @zhw. True. But OP should learn L'Hôpital's rule, it works in cases where the limit is not the definition of a derivative. – Oriol Jun 16 '16 at 20:22
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    But I would say no one should never use L'Hopital for this limit, or its notorious cousin, $\lim_{x\to 0} (\sin x)/x.$ Doing so means using what you are trying to prove in the proof! – zhw. Jun 16 '16 at 20:24