# Explain how to solve this trigonometric limit without L'Hôpital's rule?

In my previous class our professor let us the following limit:

$$\lim_{x\to0} \frac{\tan(x)-\sin(x)}{x-\sin(x)}$$

He solved it by applying L'Hôpital's rule as follow:

$\lim_{x\to0} \frac{\sec^2(x) - \cos(x)}{1-\cos(x)} = \lim_{x\to0} \frac{\cos^2(x) + \cos(x) + 1}{\cos^2(x)} = \frac{3}{1} = 3$

He only wrote that in the blackboard and then told us to solve it but without using L'Hôpital's rule. So I proceeded in this way:

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

From there I multiply by its conjugate $(1 - \cos(x))$ but then I get more confused:

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

Can someone give me a better advice on how to get the right result.

• Are you aware about Taylor's expansion? – user Apr 1 '18 at 20:58
• I've never heard about Taylor's expansion, even out professor has never told anything about it. But be sure, I'll do a depth research on it (as I am doing right now) to learn more from it. Thanks for your advice. – user547901 Apr 1 '18 at 21:06
• it is the most powerful tool to solve lamost any kind of limit. – user Apr 1 '18 at 21:07
• I'll keep in mind. Thanks again, I'm 16 (Being in the high school is the reason of I've never heard about Taylor's expansion I think) and really interested on learn new ways to solve math problems. – user547901 Apr 1 '18 at 21:11

HINT

By Taylor's expansion

$$\frac{\tan(x)-\sin(x)}{x-\sin(x)}= \frac{x+\frac{x^3}3-x+\frac{x^3}6+o(x^3)}{x-x+\frac{x^3}6+o(x^3)}=\frac{\frac{x^3}2+o(x^3)}{\frac{x^3}6+o(x^3)}=\frac{\frac{1}2+o(1)}{\frac{1}6+o(1)}\to 3$$

• The only question is if J.Doe knows Taylor's expansion. :-) – Przemysław Scherwentke Apr 1 '18 at 20:57
• @PrzemysławScherwentke Good observation! – user Apr 1 '18 at 20:58
• @PrzemysławScherwentke You're right) But I'm investigating about it just right now. This approach looks interesting. – user547901 Apr 1 '18 at 21:00
• @J.Doe I don't know if there is a trickto avoid taylor's expansion (or l'Hospital) since the game is on the third order. Of course we can't by standard limits which works when first order is involved (or second in some case). – user Apr 1 '18 at 21:04
• @J.Doe Have you got, after Langrange theorem, its extension: Cauchy theorem? If yes, maybe this is expected by your professor. – Przemysław Scherwentke Apr 1 '18 at 21:09

$$\dfrac{\tan x-\sin x}{x-\sin x}=\dfrac1{\cos x}\cdot\dfrac{\sin x}x\cdot\dfrac{1-\cos x}{x^2}\cdot\dfrac1{\dfrac{x-\sin x}{x^3}}$$

$\dfrac{1-\cos x}{x^2}=\left(\dfrac{\sin x}x\right)^2\cdot\dfrac1{1+\cos x}$

Finally for $$\dfrac{x-\sin x}{x^3}$$ use Are all limits solvable without L'Hôpital Rule or Series Expansion