# A limit without invoking L'Hopital: $\lim_{x \to 0} \frac{x \cos x - \sin x}{x^2}$

The following limit

$$\ell=\lim_{x \rightarrow 0} \frac{x \cos x - \sin x}{x^2}$$

is a nice candidate for L'Hopital's Rule. This was given at a school before L'Hopital's Rule was covered. I wonder how we can skip the rule and use basic limits such as:

$$\lim_{x \rightarrow 0} \frac{\sin x}{x} \quad , \quad \lim_{x \rightarrow 0} \frac{\cos x -1}{x^2}$$

• Can use Taylor polynomials. Nov 23, 2016 at 10:57
• Don't know that at school. Sorry ! It is high school. Nov 23, 2016 at 10:58
• Was integration covered yet? (Including the fundamental theorem of calculus) Nov 23, 2016 at 11:07
• @Orejano : According to this link en.m.wikipedia.org/wiki/L'Hôpital's_rule both are acceptable Nov 23, 2016 at 14:18
• For $0 < x < \pi$ (it suffices to look at that by parity), we have \begin{aligned}0 &< \frac{\sin x - x\cos x}{x^2} = \frac{1}{x^2} \int_0^x t\sin t\,dt \\&= \int_0^x \frac{t\sin t}{x^2}\,dt \leqslant \int_0^x \frac{t\sin t}{t^2}\,dt\\&= \int_0^x \frac{\sin t}{t}\,dt \leqslant \int_0^x 1\,dt = x,\end{aligned} since $t\sin t \geqslant 0$ on $[0,\pi]$, and $\dfrac{\sin t}{t} \leqslant 1$. Nov 23, 2016 at 14:28

From the geometric proof of $\frac{\sin x}{x} \to 1$ as $x\to0$ we know $\cos x<\frac{\sin x}{x} <1$ near $0$. Since $\frac{x\cos x -\sin x}{x^2} = \frac{\cos x -\frac{\sin x}{x}}{x}$, we see that $$0=\lim_{x\to0}\frac{\cos x -1}{x} \le \ell \le \lim_{x\to0} \frac{\cos x-\cos x}{x}=0,$$so $\ell=0$.

• The best answer. But I would speak of $\cos x < \frac{\sin x}{x} < 1$ as a reformulation of the very basic $\sin x < x < \tan x$.
– MooS
Nov 23, 2016 at 11:49
• @MooS: Thanks! The "from the geometric proof" bit was indeed referring to that, though implicitly. Nov 24, 2016 at 20:06

We have,

$$\lim_{x \to 0} \dfrac{x\cos x - \sin x}{x^2} = \lim_{x \to 0} \dfrac{\cos x -1}{x} + \lim_{x \to 0}\dfrac{x - \sin x}{x^2}$$

$$= -2\lim_{x \to 0} \dfrac{\sin^2 \left(\frac{x}{2}\right)}{x} + \lim_{x \to 0}\dfrac{x - \sin x}{x^2}$$

The first limit is zero since $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1$, and,

$$0 \leq \lim_{x \to 0}\dfrac{x - \sin x}{x^2} \leq \lim_{x \to 0}\dfrac{\tan x - \sin x}{x^2}$$

But,

$$\lim_{x \to 0}\dfrac{\tan x - \sin x}{x^2} = \lim_{x \to 0} \ \left( \sin x \times \dfrac{1-\cos x}{x^2 \cos x} \right) = \lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0$$

Thus, by the Squeeze Theorem,

$$\lim_{x \to 0} \dfrac{x\cos x - \sin x}{x^2} =0$$

The function $\frac{x\cos(x)-\sin(x)}{x^2}$ is odd, so we only need to look at $0\lt x\lt\frac\pi2$. As shown in this answer, $0\le\sin(x)\le x\le\tan(x)$. Furthermore, $x\cos(x)-\sin(x)=(x-\tan(x))\cos(x)\le0$. So we have \begin{align} 0 &\ge\color{#C00000}{\frac{x\cos(x)-\sin(x)}{x^2}}\\ &\ge\frac{\sin(x)(\cos(x)-1)}{x^2}\\ &=-\frac{\sin(x)}x\frac{1-\cos^2(x)}{x(1+\cos(x))}\\ &=-\frac{\sin^3(x)}{x^2(1+\cos(x))} \end{align} By the Squeeze Theorem, we have $$\lim_{x\to0}\frac{x\cos(x)-\sin(x)}{x^2}=0$$

• But how do you prove $\sin x\le x\le \tan x$?
– user170039
Nov 24, 2016 at 8:15
A direct way! It was fun to find it. First, we note $$\ell=\lim_{x\to0}\frac{\cos x-\frac{\sin x}{x}}{x}=\lim_{x\to0}\frac{\cos x -1}{x}+\lim_{x\to0}\frac{x-\sin x}{x^2}=\lim_{x\to0}\frac{x-\sin x}{x^2}.$$Then we try to get information from a very similar limit: \begin{align}\lim_{x\to0}\frac{\sin x\cos x-x}{x^2}=\lim_{x\to0}\frac{\sin(2x)/2-x}{x^2}&=\lim_{x\to0}\frac{\sin(2x)-2x}{2x^2} \\ &=\lim_{x\to0}\frac{\sin x-x}{\frac12x^2}\\&=-2\ell.\end{align}Now, we may also see $$\frac{x-\sin x}{x^2}=\frac{1-\frac{\sin x}{x}}{x}$$ which means $\ell=-f'(0)$, if it exists, where $f(x)=\frac{\sin x}{x}$ for $x\ne0$, $f(0)=1$. We can't tell whether $f'(0)$ exists, but the parity of $f$ ensures that if it does, it must be $0$; in particular, it can't be $+\infty$ nor $-\infty$. Therefore, what we found above implies $$\ell=\lim_{x\to0}\frac{x-x\cos x+\sin x-\sin x\cos x}{x^2}=\lim_{x\to0}(x+\sin x)\frac{1-\cos x}{x^2}=0.$$
Is it compulsory to exploit $\lim_\limits{x\to 0} \dfrac{\sin x}{x}$ and $\lim_\limits{x\to 0} \dfrac{\cos x -1}{x^2}$?
If not, why don't you just expand it, like: $$\lim_\limits{x\to 0} \frac{ x \cos x -\sin x }{ x^2 } =\lim_\limits{x\to 0} \frac{ x [1 -x^2/2 +\mathscr{O}(x^4)] -[x -x^3/6 +\mathscr{O}(x^5)] }{x^2} =\lim_\limits{x\to 0} \frac{x -x^3/2 -x +x^3/6 +\mathscr{O}(x^5)}{x^2} =\lim_\limits{x\to 0} \frac{-x^3/3}{x^2} =\lim_\limits{x\to 0} -\frac{x}{3} =0$$