How do you calculate this limit without using L'Hopital or Taylor?

I need to calculate this limit $$\lim_{x\rightarrow 0}\frac{xe^x-(e^x-1)}{x^2}$$ but only with elementary methods, so no L'Hopital/Taylor. I've done quite a bit of manipulation but nothing seems to work, could you give me a hint?

Solution 1:

Hint: Does the following help? $$\frac{xe^x-(e^x-1)}{x^2}=\frac{e^x(x-1)+1}{(x-1)(x+1)+1}=\frac{e^x+\frac{1}{x-1}}{x+1+\frac{1}{x-1}}$$

Solution 2:

$$\frac{xe^x-(e^x-1)}{x^2}=\frac{d}{dx}\left(\frac{e^x-1}{x}\right)$$ so to find you limit is equivalent to finding a second-order expansion for the exponential function in a neighbourhood of the origin. It is well-known that the exponential function is a convex function, and a fixed point for the operator $\frac{d}{dx}$, hence the inequality $$ e^{x}\geq 1+x $$ can be strengthened to: $$\forall x\in[-1,1],\qquad \left|\frac{e^x-1}{x}-\left(1+\frac{x}{2}\right)\right|\leq\frac{x^2}{4}$$ and that is enough to prove that the wanted limit equals $\color{red}{\large\frac{1}{2}}$.

Solution 3:

Clearly we have $$\frac{xe^{x} - (e^{x} - 1)}{x^{2}} = \frac{xe^{x} - x - (e^{x} - 1 - x)}{x^{2}} = \frac{e^{x} - 1}{x} - \frac{e^{x} - 1 - x}{x^{2}}\tag{1}$$ and we know that $(e^{x} - 1)/x \to 1$ as $x \to 0$ hence it follows from the equation $(1)$ that our job is done if we can calculate the limit $$\lim_{x \to 0}\frac{e^{x} - 1 - x}{x^{2}} = L\tag{2}$$ and the answer to original question would be $1 - L$.

The limit $L$ can be easily (very very easily) calculated using L'Hospital's Rule or Taylor series. But since these methods are forbidden we need to invoke some definition of $e^{x}$. The simplest approach seems to be to use the defining equation $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{3}$$ Let's first handle the case when $x \to 0^{+}$. By the binomial theorem we have $$\left(1 + \frac{x}{n}\right)^{n} = 1 + x + \dfrac{1 - \dfrac{1}{n}}{2!}x^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}x^{3} + \cdots$$ and hence \begin{align} L &= \lim_{x \to 0}\frac{e^{x} - 1 - x}{x^{2}}\notag\\ &= \lim_{x \to 0}\lim_{n \to \infty}\dfrac{\left(1 + \dfrac{x}{n}\right)^{n} - 1 - x}{x^{2}}\notag\\ &= \lim_{x \to 0}\lim_{n \to \infty}\dfrac{1 - \dfrac{1}{n}}{2!} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}x + \cdots\notag\\ &= \lim_{x \to 0}\lim_{n \to \infty}\phi(x, n)\text{ (say)}\tag{4} \end{align} If $x \to 0^{+}$ then we can see that $$\frac{1}{2} - \frac{1}{2n}\leq \phi(x, n) \leq \frac{1}{2} + \frac{x}{3!} + \frac{x^{2}}{4!}\cdots \leq \frac{1}{2} + \frac{x}{2\cdot 3} + \frac{x^{2}}{2\cdot 3\cdot 3} + \cdots = \frac{1}{2} + \frac{x}{6 - 2x}\tag{5}$$ for $0 < x < 3$. Further note that $\phi(x, n)$ is increasing as $n$ increases and by above equation $(5)$ it is bounded above. Hence $\lim_{n \to \infty}\phi(x, n) = \phi(x)$ exists for $0 < x < 3$. From equation $(5)$ it follows that $$\frac{1}{2} \leq \phi(x) \leq \frac{1}{2} + \frac{x}{6 - 2x}$$ and then by Squeeze Theorem $\phi(x) \to 1/2$ as $x \to 0^{+}$. It now follows from equation $(4)$ that $L = 1/2$ (as far $x \to 0^{+}$ is concerned).

The case $x \to 0^{-}$ is easy (try it and it will surprise you!). Thus $L = 1/2$ and desired limit $(1 - L) = 1/2$.