Limit of $\frac{\tan(x)-x}{x^3}$ as $x$ approaches $0$ without L'Hospital's Rule

Question

I am trying to find the limit of $\frac{\tan(x)-x}{x^3}$ as $x$ approaches $0$. I know that this can be found by using L'Hospital's Rule 3 times. Is there a way to solve this problem without using L'Hospital's Rule?

Please do not use Taylor series; I consider this to be an equivalent method. I have noticed that the required number of applications of L'Hospital's Rule is precisely the order of the first non-zero derivative, which I think is essentially because a product is $0$ if and only if at least one factor is $0$.

@user1551 I am only interested in the specific limit in the title — cpiegore, Feb 14 '16 at 18:56
How about breaking up the limit so you have ${tanx\over x^3} - {1/x^2}$? Then multiply above and below by $sin^3x$ — Morgormir, Feb 14 '16 at 19:40
@cpiegore your limit is the first limit solved in the link suggested by user1551. Go there and you'll find the answer. — Giovanni Resta, Feb 14 '16 at 20:22

score 1 · Accepted Answer · answered Feb 14 '16 at 19:43

1

you can simplify $$\frac{\tan x - x}{x^3}= \frac{\sin x - x \cos x}{x^3\cos x } = \frac{x - \frac{x^3}{6}+\cdots - x\left(1 - \frac{x^2}2+\cdots\right)}{x^3} = \frac 13 \text{ as } x \to 0. $$

answered Feb 14 '16 at 19:43

abel

29,170

Limit of $\frac{\tan(x)-x}{x^3}$ as $x$ approaches $0$ without L'Hospital's Rule

1 Answers1