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Any ideas on how to calculate the limit of $(e^x -1)/{x}$ as $x$ goes to zero without applying L'Hôpital's rule?

Martin Sleziak
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Spinning Turntable
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  • Do you know what is the derivative of $x\mapsto e^x$? – Git Gud Dec 16 '14 at 12:47
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    $x\leq e^x-1 \leq xe^x$ an squeeze theorem – Surb Dec 16 '14 at 12:48
  • the basic fact is $e$ is defined as either $\lim_{n\to\infty}(1+1/n)^n$ or $\sum_{k=0}^{\infty}k!^{-1}$ – RE60K Dec 16 '14 at 14:00
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    Some of the answers in other posts might answer your question. For example, see these questions: http://math.stackexchange.com/questions/152605/proving-that-lim-limits-x-to-0-fracex-1x-1, http://math.stackexchange.com/questions/403876/show-that-lim-limits-x-to0-fracex-1x-1 – Martin Sleziak Dec 17 '14 at 13:48

4 Answers4

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Using Bernoulli's Inequality, for all $x$ so that $|x|\le n$, $$ 1+x\le\left(1+\frac xn\right)^n\tag{1} $$ Therefore, letting $n\to\infty$, we get for all $x$, $$ 1+x\le e^x\tag{2} $$ Furthermore, for $|x|\lt1$, $$ 1-x\le e^{-x}\implies\frac1{1-x}\ge e^x\tag{3} $$ Thus, subtracting $1$ from $(2)$ and $(3)$ gives $$ x\le e^x-1\le\frac x{1-x}\tag{4} $$ Since we are looking for the limit as $x\to0$, assume that $|x|\lt1$. Whether $x$ is positive or negative, $(4)$ says that $$ \frac{e^x-1}{x}\text{ is between $1$ and }\frac1{1-x}\tag{5} $$ Therefore, by the Squeeze Theorem, we get $$ \lim_{x\to0}\frac{e^x-1}{x}=1\tag{6} $$

robjohn
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I don't know if this is really "without" Hôpitals rule for you but if you are allowed to use $$ \exp(x)=\sum_{k=0}^\infty \frac{x^k}{k!} $$ the limit is straightforward since this sum converges locally uniformly.

frog
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Observe that when $x$ is very small near $0$, we have:

If $x > 0 \Rightarrow 1+x < e^x < 1 + x + 2x^2 \to 1 < \dfrac{e^x-1}{x} < 1 + 2x$, and letting $x \to 0^{+}$, we have: $\displaystyle \lim_{x \to 0^{+}} \dfrac{e^x-1}{x} = 1 \tag{1}$,

and for $x < 0 \Rightarrow 1 > \dfrac{e^x-1}{x} > 1 + 2x$, and letting $x \to 0^{-}$, we have: $\displaystyle \lim_{x\to 0^{-}} \dfrac{e^x-1}{x} = 1 \tag{2}$.

$(1),(2) \Rightarrow \displaystyle \lim_{x \to 0} \dfrac{e^x-1}{x} = 1$.

DeepSea
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$$e^x=1+x+1/2x^2+\cdots$$

$${e^x-1\over x}={x+x^2/2+\cdots\over x}=1+\frac12x+\cdots$$

$$\lim_{x\to0}{e^x-1\over x}=\lim_{x\to0}1+\frac12x+\cdots=1.$$

Suzu Hirose
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