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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:

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

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    $\begingroup$ Can use Taylor polynomials. $\endgroup$ – Teddy Nov 23 '16 at 10:57
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    $\begingroup$ Don't know that at school. Sorry ! It is high school. $\endgroup$ – Tolaso Nov 23 '16 at 10:58
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    $\begingroup$ Was integration covered yet? (Including the fundamental theorem of calculus) $\endgroup$ – Daniel Fischer Nov 23 '16 at 11:07
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    $\begingroup$ @Orejano : According to this link en.m.wikipedia.org/wiki/L'Hôpital's_rule both are acceptable $\endgroup$ – Tolaso Nov 23 '16 at 14:18
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    $\begingroup$ 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 $\dfrac{\sin t}{t} \leqslant 1$. $\endgroup$ – Daniel Fischer Nov 23 '16 at 14:28
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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,

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

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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 $\ell=0$.

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    $\begingroup$ The best answer. But I would speak of $\cos x < \frac{\sin x}{x} < 1$ as a reformulation of the very basic $\sin x < x < \tan x$. $\endgroup$ – MooS Nov 23 '16 at 11:49
  • $\begingroup$ @MooS: Thanks! The "from the geometric proof" bit was indeed referring to that, though implicitly. $\endgroup$ – Vincenzo Oliva Nov 24 '16 at 20:06
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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 $$ \lim_{x\to0}\frac{x\cos(x)-\sin(x)}{x^2}=0 $$

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  • $\begingroup$ But how do you prove $\sin x\le x\le \tan x$? $\endgroup$ – user 170039 Nov 24 '16 at 8:15
  • $\begingroup$ @user170039: this answer should provide a geometric justification. I have added a link above. $\endgroup$ – robjohn Nov 24 '16 at 8:53
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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 $$\ell=\lim_{x\to0}\frac{x-x\cos x+\sin x-\sin x\cos x}{x^2}=\lim_{x\to0}(x+\sin x)\frac{1-\cos x}{x^2}=0. $$

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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: $$ \lim_\limits{x\to 0} \frac{ x \cos x -\sin x }{ x^2 } =\lim_\limits{x\to 0} \frac{ x [1 -x^2/2 +\mathscr{O}(x^4)] -[x -x^3/6 +\mathscr{O}(x^5)] }{x^2} =\lim_\limits{x\to 0} \frac{x -x^3/2 -x +x^3/6 +\mathscr{O}(x^5)}{x^2} =\lim_\limits{x\to 0} \frac{-x^3/3}{x^2} =\lim_\limits{x\to 0} -\frac{x}{3} =0 $$

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  • $\begingroup$ If you had seen the comments students at High School do not know Taylor expansion. So your solution is not valid. Anyway , thanks for writing that down. $\endgroup$ – Tolaso Nov 26 '16 at 9:33
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    $\begingroup$ Well, I did not see that comment only until now! Never mind, my fault. btw, I think there is nothing wrong to introduce Maclaurin expansion early, even to high school students you can tell them, the series is just close to sin and cos, which you will learn why someday! All they have to know, is that rhs and lhs are about equal. To me, it is more difficult to exploit strange limit tricks than just to apply Maclaurin. If you think it's bad that they just memorize Maclaurin, draw graphs to give them intuition. $\endgroup$ – Violapterin Nov 26 '16 at 9:41
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    $\begingroup$ Taylor expansions and L'Hôpital rule are essentially equivalent, even if not necessarily in practice. Moreover, drawing graphs does note give students intuition, it just verifies the "dogma". I think no-L'Hôpital (and no-Taylor) exercises are a good tool to develop creativity and reasoning skills of students. $\endgroup$ – Vincenzo Oliva Nov 26 '16 at 10:29

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