# 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?

• What counts as "elementary methods"? (Can't get more elementary than the power series expansion from my point, since that's how $e^x$ was defined in our course.) – Daniel Fischer Dec 8 '15 at 9:50
• Algebraic manipulation of the expression and simple well-known limits like $\frac{e^x-1}{x}$. Basically any technique to solve limits that is learnt before derivatives. – Nicol Dec 8 '15 at 9:55
• That would depend on location (and time) of study. For me, that would pretty much limit you to geometric series, since derivatives came rather early (after the theorems about limits and continuous functions were proved). – Daniel Fischer Dec 8 '15 at 9:57

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

• I don't understand how this will help finding the limit, as we still in an indeterminante form! – Nizar Dec 8 '15 at 9:45
• Maybe OP does, he asked for a hint... – Eric S. Dec 8 '15 at 9:49
• I don't see how the hint is usefull as well. As the limit is to zero, we have $\frac{\infty}{\infty}$ – Tomas Dec 8 '15 at 9:57
• Since the numerator and denominator look alike in this form, the squeeze theorem becomes particularly useful – Kelvin Soh Dec 8 '15 at 10:00
• Could you please elaborate on your hint? I am unable to see how this translates to the limit $1/2$. – Paramanand Singh Dec 9 '15 at 4:13

$$\frac{xe^x-(e^x-1)}{x^2}=\frac{d}{dx}\left(\frac{e^x-1}{x}\right)$$ 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 $\color{red}{\large\frac{1}{2}}$.

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 $(1 - L) = 1/2$.