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

Question

$$\lim \limits _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$$

I tried changing separating the terms and converting to $\tan x$ but I got stuck. A little help would be helpful.

Answer 1

$$\lim_{x\to0}\frac{x\cos(x)-\sin(x)}{x^2\sin(x)}=$$

---

Apply l'Hôpital's rule, to get:

• $$\frac{\text{d}}{\text{d}x}\left(x\cos(x)-\sin(x)\right)=-x\sin(x)$$
• $$\frac{\text{d}}{\text{d}x}\left(x^2\sin(x)\right)=x^2\cos(x)+2x\sin(x)$$

---

$$\lim_{x\to0}-\frac{\sin(x)}{x\cos(x)+2\sin(x)}=-\lim_{x\to0}\frac{\sin(x)}{x\cos(x)+2\sin(x)}=$$

---

Apply l'Hôpital's rule again, to get:

• $$\frac{\text{d}}{\text{d}x}\left(\sin(x)\right)=\cos(x)$$
• $$\frac{\text{d}}{\text{d}x}\left(x\cos(x)+2\sin(x)\right)=3\cos(x)-x\sin(x)$$

---

$$-\lim_{x\to0}-\frac{\cos(x)}{x\sin(x)-3\cos(x)}=\lim_{x\to0}\frac{\cos(x)}{x\sin(x)-3\cos(x)}=\frac{\cos(0)}{0\sin(0)-3\cos(0)}=-\frac{1}{3}$$

Answer 2

Use taylor expansions

$cos(x) = 1 - \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}+...$

$x.cos(x) = x - \frac{x^3}{2!}+ \frac{x^5}{4!}- \frac{x^7}{6!}+...$

$sin(x) = x - \frac{x^3}{3!}+ \frac{x^5}{5!}- \frac{x^7}{7!}+...$

(1) $x.cos(x) - sin(x) = x^3.(- \frac{1}{2!}+\frac{1}{3!}) + x^5.(a_5) + ...$

$x^2.sin(x) = x^3 - \frac{x^5}{3!}+ \frac{x^7}{5!}- \frac{x^9}{7!}+...$

(2) $x^2.sin(x) = x^3(1 - \frac{x^2}{3!}+ \frac{x^4}{5!}- \frac{x^6}{7!}+...)$

Thus;

$\frac{x.cos(x) - sin(x)}{x^2.sin(x)} = \frac{(- \frac{1}{2!}+\frac{1}{3!}) + x^2.a_5 + ...}{(1 + x^2.b_5 + ...)}$

$\implies \lim_{x \to 0} \frac{x.cos(x) - sin(x)}{x^2.sin(x)} = - \frac{1}{2!}+\frac{1}{3!} = - \frac{1}{3} $

Answer 3

$$\dfrac{x\cos x-\sin x}{x^2\sin x}=\cos x\cdot\dfrac{x-\tan x}{x^3}\cdot\dfrac x{\sin x}$$

As $F=\text{lim}_{x\to0}\dfrac{x-\tan x}{x^3}$ is of the form $\dfrac00$

Apply L'Hospital's Rule to find $F=\text{lim}_{x\to0}\dfrac{1-\sec^2x}{3x^2}=-\dfrac13\cdot\left(\text{lim}_{x\to0}\dfrac{\sin x}x\right)^2\dfrac1{\text{lim}_{x\to0}\cos^2x}=?$

Alternative use Series Expansion, $$\tan x=x+\dfrac{x^3}3+O(x^5)$$