Intereting Posts

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How can we prove that

$$\lim_{x\rightarrow 0}\cfrac{e^x-1-x}{x^2}=\cfrac{1}{2}$$ Without using L’hopital rule, and Taylor expansions?

Thanks

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Use the limit laws and the binomial theorem: you have $$\frac{e^x – (1+x)}{x^2} = \lim_{n \to \infty} \left( \frac{ (1+ \frac xn)^n – (1+x)}{x^2} \right) = \lim_{n \to \infty} \sum_{k=2}^n \binom nk \frac{x^{k-2}}{n^k} \\ = \frac 12 + x \left( \lim_{n \to \infty} \sum_{k=3}^n \binom nk \frac{x^{k-3}}{n^k} \right)$$

**provided that the limit** $ \displaystyle e^x = \lim_{n \to \infty} \left(1 + \frac xn \right)^n$ **is assumed to exist**.

As a by-product of this computation you get that $$\lim_{n \to \infty} \sum_{k=3}^n \binom nk \frac{x^{k-3}}{n^k}$$ exists too. With $x=1$ this implies $$\sup_n \sum_{k=3}^n \binom nk \frac{1}{n^k} < \infty$$

and consequently if $|x| \le 1$ then $$\sup_{n} \left| \sum_{k=3}^n \binom nk \frac{x^{k-3}}{n^k} \right| \le \sup_{n} \sum_{k=3}^n \binom nk \frac{|x^{k-3}|}{n^k} \le \sup_n \sum_{k=3}^n \binom nk \frac{1}{n^k} < \infty.$$

So, if $|x| \le 1$ then $$\left| \frac{e^x – (1+x)}{x^2} – \frac 12 \right| \le |x| \sup_n \sum_{k=3}^n \binom nk \frac{1}{n^k}.$$

Now let $x \to 0$.

For every natural $n\ge 2$ we have

$$\lim_{x\to 0}\frac{\left(1+\frac xn\right)^n-x-1}{x^2}=\lim_{x\to0}\frac{1+n\frac xn+\binom n2\frac{x^2}{n^2}+x^3P(x)-x-1}{x^2}=\frac{n-1}{2n}$$

where $P(x)$ is a polynomial.

This alone does not imply that your limit is $1/2$. We need to assume that the function

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

is defined in some interval $I$ around $0$ (that is, that the limit exists for every $x\in I$) and continuous.

Are derivatives allowed? Assuming the existence of $$L = \lim_{x \to 0} \frac{e^x – 1 – x}{x^2} = \lim_{x \to 0} \frac{\left(\frac{e^x – 1}x\right) – 1}x$$ it is equal to the derivative of the (continuous) function $$f(x) = \begin{cases} \frac{e^x – 1}x & x \neq 0 \\ 1 & x = 0 \end{cases}$$ at $x = 0$. Thus, we have (formally) $$L = \lim_{t \to 0}\left.\frac{d}{dx}\left(\frac{e^x – 1}x\right)\right|_{x = t} = \lim_{t \to 0} \left(\frac{e^t}t – \frac{e^t – 1}{t^2}\right) = \lim_{t \to 0} \left(\frac{e^t – 1}t – \frac{e^t – 1 – t}{t^2}\right) = 1 – L$$

We have that $f(x)=e^x$ is an increasing convex function, hence for any $\varepsilon>0$ there is a neighbourhood of the origin $U$ where the inequality

$$ (1-\varepsilon)x\leq e^x-1 \leq (1+\varepsilon)x \tag{1}$$

holds. If we consider some $z\in U$ and integrate every term of $(1)$ over $(0,z)$ we get

$$ (1-\varepsilon)\frac{z^2}{2}\leq e^z-1-z \leq (1+\varepsilon)\frac{z^2}{2}\tag{2} $$

hence

$$ \liminf_{z\to 0}\frac{e^z-1-z}{z^2}\geq\frac{1-\varepsilon}{2},\qquad \limsup_{z\to 0}\frac{e^z-1-z}{z^2}\leq \frac{1+\varepsilon}{2} \tag{3}$$

and since $\varepsilon$ is arbitrary, the claim follows.

I guess we’re allowed limits here, so take $e^x = \lim_{k\rightarrow \infty}(1+x/k)^k$. So for large $k$, your limit should be about the same as $$\lim_{x\rightarrow 0} \frac{(1+x/k)^k – 1 – x}{x^2} = \lim_{x\rightarrow 0} \frac{1 + {k\choose 1}\frac{x}{k} + {k \choose 2}\frac{x^2}{k^2} + \mbox{HOT} – 1 – x}{x^2} =\lim_{x\rightarrow 0} \frac{{k \choose 2}\frac{x^2}{k^2} + \mbox{HOT}}{x^2} =\frac{1}{2}.$$

If we can figure out a way to swap the two limits, we’ll have a rigorous proof.

If you allow a few basic calculus results into the game, then note

$$e^x-x – 1 = \int_0^x (e^t-1)\, dt = \int_0^x [(e^t-1)/t]t\, dt.$$

Now $(e^t-1)/t \to 1$ as $t\to 0,$ simply because the derivative of $e^x$ is $1$ at $0.$ That implies the last integral above $\approx \int_0^x t\, dt = x^2/2$ and that will lead to the result. So a few details missing, but it’s not too hard to fill them in.

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