Find limits of $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor Expansion.

Question

Find the limit $\displaystyle \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor expansion.

My Try

$\displaystyle =\lim _{x \to 0} \frac {\cos x - \frac{\sin x}{x}} {x \sin x}$

$=\frac {\displaystyle\lim _{x \to 0}\cos x - \lim _{x \to 0}\frac{\sin x}{x}} {\displaystyle\lim _{x \to 0}x \sin x}$

$=\frac{1-1}{0}$

But still I end up with $\frac00$

Any hint for me to proceed would be highly appreciated.

P.S: I did some background check on this question on mathstack and found they have solved this with l'Hopital's rule and the answer seems to be $\frac{-1}{3}$.

What is $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$?

Answer 1

Famously $\lim_{x\to0}\frac{\sin x}{x}=1$ can be proved without such techniques, and implies $\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$. With a little more effort (e.g. by approximating a circular arc as a parabola), you can also show $\lim_{x\to0}\frac{x-\sin x}{x^3}=\tfrac16$. So $\lim_{x\to0}\frac{x-\sin x}{x^2\sin x}=\tfrac16$ and$$\lim_{x\to0}\left(\frac{\cos x-1}{x\sin x}+\frac{x-\sin x}{x^2\sin x}\right)=-\frac12+\frac16=-\frac13.$$

Answer 2

We have

$$\frac {x \cos x - \sin x} {x^2 \sin x}=\frac{\cos x}{\frac{\sin x}x}\cdot\frac{x-\tan x}{x^3}\to1\cdot \left(-\frac13\right)=-\frac13$$

using

• How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
• Are all limits solvable without L'Hôpital Rule or Series Expansion

Answer 3

While searching more easy way, let me suggest one possible way to solve main difficult part of suggested limit. I change denumerator to $x^3$, for simplicity, as it's equivalent $x^2\sin x$

Suppose we know existence of limit. Then $$L=\lim_{x\to0}\frac{x-\sin x}{x^3} = \lim_{x\to0}\frac{x-3\sin \frac{x}{3}+4 \sin^3 \frac{x}{3}}{x^3}=\\ =\lim_{x\to0}\left(3\frac{\frac{x}{3} - \sin \frac{x}{3}}{x^3} + \frac{4 \sin^3 \frac{x}{3}}{x^3}\right) =\lim_{x\to0}\left(\frac{\frac{x}{3} - \sin \frac{x}{3}}{9\left(\frac{x}{3}\right)^3} + \frac{4 \sin^3 \frac{x}{3}}{x^3}\right)=\frac{L}{9}+\frac{4}{27}$$ From obtained equation $L=\frac{1}{6}$

Answer 4

Denote $L$ the existing limit. Then, express it as

\begin{align} L=\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} &= \lim _{x \to 0}\frac {x (2\cos^2\frac x2 -1) -2 \sin \frac x2 \cos\frac x2} {2x^2 \sin \frac x2 \cos\frac x2}\\ &= \lim _{x \to 0} \frac {x (\cos^2\frac x2-1) +2 \cos\frac x2(\frac x2\cos \frac x2- \sin\frac x2)} {2x^2 \sin \frac x2 \cos\frac x2}\\ &= - \lim _{x \to 0}\frac{\sin\frac x2}{\frac x2} \frac1{4\cos\frac x2} + \lim _{x \to 0} \frac{\frac x2\cos \frac x2- \sin\frac x2} {4(\frac x2)^2 \sin \frac x2} \\ &= -\frac14+\frac14L \end{align}

Thus, $L= -\frac13$.

Answer 5

The limit in question is equal to the second derivative of the sinc function evaluated at $0$. That is,

$$\begin{align} \lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2\sin(x)}&=\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^3}\frac{x}{\sin(x)}\\\\ &=2\lim_{h\to 0}\frac{\text{sinc}(h)-1}{h^2}\\\\ &=2\lim_{h\to 0}\frac{\sin(h)-h}{h^3}\\\\ \end{align}$$

In This Answer, I showed, without use of calculus, that the sine function satisfies the inequality

$$\sin(h)\ge h-\frac16 h^3\tag1$$

In a parallel development, one can show, without calculus, that $\sin(h)\le h-\frac16h^3+\frac1{120}h^5$. (Alternatively, integrate $(1)$ twice and use $\cos(0)=1$ and $\sin(0)=0$.)

Hence, applying the squeeze theorem, we find that

$$\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2\sin(x)}=-\frac13$$