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Here is my attempt at finding a proof that $\lim_{x \to 0}\frac{e^x-1}{x} = 1$. We need to show that $\forall \varepsilon > 0 \, \exists \delta > 0$ such that $ 0 < |x| < \delta \implies |\frac{e^x-1}{x} - 1| < \varepsilon$. We can let $\delta = \min(1,?)$ where $?$ is a function of $\varepsilon$. So $$ \left|\frac{e^x-1}{x} - 1\right| < \varepsilon$$ $$ \left|\frac{e^x-1-x}{x}\right| < \varepsilon$$ The numerator of the function can be bounded by $|x| < \delta \leq 1$: $$ |e^x-1-x| \leq |e^x| + |x| + 1$$ $$ |e^x| + |x| + 1 \leq e + 1 + 1 \leq e + 2$$ But then the $|x|$ is in the denominator so: $$ \frac{|e^x-1-x|}{|x|} \nleq e + 2$$ I am not sure what to do now or if there is a better way to approach this problem. I am new to $\epsilon$-$\delta$ proofs so any help would be greatly appreciated. I am specifically looking for an approach that does not appeal to the derivative definition, $\frac{d}{dx}e^x = e^x$, or the power series $e^x = 1 + x + x^2/2! + \cdots$, wanting to stick only with limits. I am not sure if this is possible, as I assume the definition of $e^x$ must be important to the proof. I had the idea to use something like the squeeze theorem using $\epsilon$-$\delta$, but am not sure which functions to choose.

A definition of $e$ that might work (since it is a limit and does not involve derivatives) is $$ e = \lim_{n \to \infty}\left(1 + \frac{1}{n} \right)^{n}$$ But I am not sure how this definition could be used in the $\epsilon$-$\delta$ proof.

Shuhul Mujoo
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1 Answers1

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If you like $\lim_{n\to \infty}(1+\frac{1}{n})^n$ as the definition of $e$, would taking $e^x := \lim_{n\to \infty}(1+\frac{x}{n})^n$ be acceptable? If it is, then you can deduce the limit using only the binomial theorem and the sum of a geometric series:

For $n \geq 2$, let $\epsilon_n(x) = (1+\frac{x}{n})^n-1-x$. Then $$ \begin{split} \epsilon_n(x) &= (1+\frac{x}{n})^n-1-x = \sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k}-1-x\\ &=\sum_{k=2}^n\frac{n(n-1)\ldots(n-k+1)}{n^k}.\frac{x^k}{k!}\\ \end{split} $$ Now since $1/(1+1)!= 1/2^1$, it follows by induction that $1/(l+1)!\leq 1/2^l$, hence find that for $n\geq 2$ and $|x|<2$ $$ \begin{split} |\epsilon_n(x)| &\leq \sum_{k=2}^n\frac{n}{n}\frac{(n-1)}{n}\ldots \frac{(n-k+1)}{n}\frac{|x|^k}{k!}\\ &\leq \sum_{k=2}^n \frac{|x|^k}{k!} =|x|\sum_{l=1}^{n-1}\frac{|x|^{l}}{(l+1)!} \leq |x|\sum_{l=1}^{n-1} (\frac{|x|}{2})^l\\ &\leq |x|\sum_{l=1}^{\infty}(\frac{|x|}{2})^l = |x|(\frac{|x|/2}{1-|x|/2})=\frac{|x|^2}{2-|x|} \end{split} $$ But by definition, $\epsilon_n(x) \to e^x-1-x$ as $n \to \infty$, hence it follows that $|e^x-1-x|\leq \frac{|x|^2}{2-|x|}$, and hence $$ |\frac{e^x-1}{x}-1|\leq \frac{|x|}{2-|x|} \to 0 \text{ as } x\to 0, $$ hence $\lim_{x\to 0} \frac{e^x-1}{x} = 1$ as required.

krm2233
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  • Isn't $k!\ge 2^k$ valid only for $k \ge 4$? Moreover, can you explain why is $\sum_{k=2}^n \frac{|x|^k}{k!} \le |x| \sum_{k=1}^\infty \frac{|x|^k}{2^k}$? I get it that you have estimated the finite sum with the series and you used the inequality $k! \ge 2^k$, but I don't get why there is a factor $|x|$ that multiplies the series. – ZaWarudo Jun 23 '23 at 12:47
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    I've edited the answer to try and clarify the estimate -- you're right that $k!\geq 2^k$ for $k\geq 4$, but you can use $(k+1)! \geq 2^k$ instead, which is true for all $k\geq 1$ (same inductive proof, but you get the base case $(1+1)^1=2^1$). The factor of $|x|$ is what gives you the $(l+1)!$ -- the sum starts from $k=2$, so taking $l+1=k$, the sum becomes one that starts at $l=1$, with a $|x|^{l+1}$ out of which you can then take a factor of $|x|$. – krm2233 Jun 23 '23 at 18:31
  • Thanks for clarifying :D – ZaWarudo Jun 23 '23 at 19:31