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Is there any way to evaluate $$\lim_{x\to0} \frac{x \cos x - \sin x}{x^2}$$ without using L'Hopital's Rule? I was trying to use some of the standard trig limits (e.g. $\lim_{x\to 0} \frac{\sin x}{x} = 1)$, etc. but couldn't figure it out.

Thank you.

Conan Wong
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6 Answers6

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Adding and subtracting $x$ in nummerator we can split the expression under limit as $$\frac{\cos x-1} {x} +\frac{x-\sin x} {x^2}$$ The first expression tends to $0$ because $(1-\cos x) /x^2\to 1/2$ and the second expression also tends to $0$ as shown below.

Since the second expression is an odd function it is sufficient to prove that the expression tends to $0$ as $x\to 0^{+}$. Next note the famous inequality $$\sin x<x<\tan x$$ for $0<x<\pi/2$. Using the above inequality we get $$0<\frac{x-\sin x} {x^2}<\frac{\tan x-\sin x} {x^2}=\tan x\cdot\frac{1-\cos x} {x^2}$$ and applying Squeeze Theorem we get the desired limit as $0$. Thus the limit in question is $0$.

Paramanand Singh
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$\begin{align} \lim_{x \to 0} \frac{x\cos(x)-\sin(x)}{x^2}\sim\\ \sim\lim_{x \to 0} \frac{\sin(x)\cos(x)-\sin(x)}{x^2}=\\ =\lim_{x \to 0} \sin(x)\frac{\cos(x)-1}{x^2}=\\ =\lim_{x \to 0} \sin(x)\frac{-2\sin^2(\frac{x}{2})}{x^2}=\\ =\lim_{x \to 0} \sin(x) \cdot\left(-\frac{1}{2}\right)\frac{\sin^2(\frac{x}{2})}{\frac{x}{2}^2}=\\ =\lim_{x \to 0} -\frac{1}{2}\sin(x)=\\ =0 \end{align}$

Note: You can easily avoid half of the computation assuming as already proved the trig limit $\lim_{x \to 0}\frac{\cos(x)-1}{x^2}=-\frac{1}{2}$

The asymptotic is justified since $\sin(x)=x+o(x^2)$

To proove it, consider: $\cos(x)<\frac{\sin(x)}{x}<1$

Thus, multiplying all for $x$ and then subtracting $x$ and dividing for $x^2$, we get:

$\frac{x\cos(x)-x}{x^2}<\frac{\sin(x)-x}{x^2}<0$

Thus, as we have showned before, the left term goes to $0$, and using the squeeze theorem, we have shown the statement

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    you can't substitute x with $\sin x$ – user Jan 20 '18 at 09:14
  • As mentioned by @gimusi your solution is wrong and the mistake is very common. You may want to fix this to avoid any downvotes. – Paramanand Singh Jan 20 '18 at 09:28
  • As showed by Paramanand you can fix your idea by using inequalities! – user Jan 20 '18 at 09:47
  • It is not necessary @gimusi. I have used the asymptotic relation $x=\frac{x}{\sin(x)}\sin(x)$~$\sin(x)$ and the fact that the numerator is bounded –  Jan 20 '18 at 09:56
  • @gimusi It is not a substitution, as I am trying to show with my comment. Maybe it would be useful anyway for me to explain the passage. What do you think? –  Jan 20 '18 at 09:59
  • @gabrielecassese Yes I think you should explain your method woth further details, as it is presented now it seems a substitution. – user Jan 20 '18 at 10:11
  • The problem with your asymptotic is that the denominator also tends to $0$ and then you need more careful analysis. Otherwise using your approach $(x-\sin x) /x^3\to 0$. – Paramanand Singh Jan 20 '18 at 10:15
  • @ParamanandSingh I have used this asymptotic extimation since it has an error o(x²), otherwise it is wrong. I didn't thought is was necessary to explain it –  Jan 20 '18 at 10:21
  • @ParamanandSingh I have explained all the passages in the answer, thought –  Jan 20 '18 at 10:34
  • @ParamanandSingh from your work here https://math.stackexchange.com/questions/1809293/taylors-theorem-with-peanos-form-of-remainder it seems that Taylor'e expansion with Peano's reminder can be shown without l'Hopital, is it correct? – user Jan 20 '18 at 10:38
  • The equation $\sin x=x+o(x^2)$ is a non-trivial result which is needs to be proved like I did in my answer. You just assume it. – Paramanand Singh Jan 20 '18 at 10:42
  • @gimusi : yes I have given one such proof there. But you can discuss it in that question instead of here. – Paramanand Singh Jan 20 '18 at 10:42
  • @ParamanandSingh I cannot proove anything –  Jan 20 '18 at 10:43
  • Well in that case you can just assume the usual Taylor expansion $\sin x =x-x^3/6+o(x^3)$ and say that your solution is based on Taylor expansions. – Paramanand Singh Jan 20 '18 at 10:45
  • Anyway I have removed my downvote based on your assumption $\sin x=x+o(x^2)$. – Paramanand Singh Jan 20 '18 at 10:46
  • @ParamanandSingh I proved it without using Taylor's theorem –  Jan 20 '18 at 11:24
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Following Paramanand Singh method by inequalities, we have that

$$\frac{x \cos x - x}{x^2}\leq\frac{x \cos x - \sin x}{x^2}\leq\frac{\tan x \cos x - \sin x}{x^2}=0$$

and since

$$\frac{x \cos x - x}{x^2}=x\frac{\cos x - 1}{x^2}\to 0\cdot -\frac12=0$$

for squeeze theorem the limit is equal to 0.

user
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I would then advise using the Taylor series expansion for $\cos x$ and $\sin x$ giving us: $$L =\lim_{x\to 0}\frac{x\cos x - \sin x} {x^2}$$ $$=\lim_{x\to 0} \frac{x\left[1-\frac{x^2}{2!} +\frac{x^4}{4!}-\ldots\right]-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots \right)} {x^2}$$ $$=0$$

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$$\begin{align}\lim_{x\to0} \frac{x \cos x - \sin x}{x^2}\\&=\lim_{x\to0} \dfrac{x\cos x-x+x-\sin x}{x^2}\\ &=\lim_{x\to0}\dfrac{x(\cos x-1)+(x-\sin x)}{x^2}\\&=\lim_{x\to0} (\dfrac{\cos x-1}{x^2})x+(\dfrac{x-\sin x}{x^2})\\&=(\dfrac{-1}{2})(0)+0\\&=0\end{align}$$

Fricul38
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$f(x)=\frac{x\cos x-\sin x}{x^2}$ is an odd function on $\mathbb{R}\setminus\{0\}$, hence if the limit exists, it is zero.
It is enough to prove that the limit as $x\to 0^+$ exists, and for such a purpose one may simply employ the classical inequalities $x-\frac{x^3}{6}\leq \sin x\leq x$ and $1-\frac{x^2}{2}\leq \cos x\leq 1$ in a right neighbourhood of the origin.

Jack D'Aurizio
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