Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor

Question

How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$

without using L'Hospital or Taylor series?

thanks :)

Answer 1

Let $L = \lim_{x \to 0} \dfrac{x - \sin(x)}{x^3}$. We then have \begin{align} L & = \underbrace{\lim_{y \to 0} \dfrac{3y - \sin(3y)}{27y^3} = \lim_{y \to 0} \dfrac{3y - 3\sin(y) + 4 \sin^3(y)}{27y^3}}_{\sin(3y) = 3 \sin(y) - 4 \sin^3(y)}\\ & = \lim_{y \to 0} \dfrac{3y - 3\sin(y)}{27 y^3} + \dfrac4{27} \lim_{y \to 0} \dfrac{\sin^3(y)}{y^3} = \dfrac{3}{27} L + \dfrac4{27} \end{align} This gives us $24L = 4 \implies L = \dfrac16$

Answer 2

Note: The original question didn't say anything about not using Taylor series. Then, after I answered this, the OP changed the question and said didn't want Taylor series either.

Another option would be using power series (aka Maclaurin series aka Taylor series centered at 0)

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$

At this point, it's pretty easy.

Answer 3

Short answer: function discussion. Long answer (details) is here.

First step. Your fraction is an even function so it is enough to consider the right limit.

Second step. Define the functions $f(x):=\sin(x)-x+\frac{1}{6}x^3$ and $g(x):=\sin(x)-x+\frac{1}{6}x^3-\frac{1}{120}x^5$ on the interval $[0, 0.1]$.

Third step. Prove that $f$ is strictly increasing and $g$ is strictly decreasing. ($f(0)=0$, it is enough $f'>0$. $f'(0)=0$, so it is enough $f''>0$ and so on. Similarly for $g$.

Forth step. From the third step $$ x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5. $$ The limit follows.

Answer 4

Assuming $f$ is sufficiently smooth, repeated application of the fundamental theorem of the calculus gives (finite Taylor expansion) $$f(x) = f(0)+f'(0)x+\frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \int_0^x (x-t)^2 \frac{(f'''(t)-f'''(0))}{2!}\, dt$$

Using the fact that $\sin' = \cos, \cos' = -\sin$, we can then expand $f(x) = \sin x$ as $$ \sin x = x -\frac{x^3}{6} + \int_0^x (x-t)^2 \frac{(-\cos t +1)}{2!}\, dt$$ Let $\epsilon>0$, and choose $\delta>0$ such that if $|t|< \delta$, then $|1-\cos t| < \epsilon$. Then, replacing $1-\cos t$ by $\epsilon$ and integrating, we have the estimate $$ \left|\sin x - x + \frac{x^3}{6} \right| \leq \epsilon \frac{|x|^3}{6}$$ If $0 < |x| < \delta$, then dividing through by $|x^3|$ gives: $$ \left| \frac{\sin x - x}{x^3} +\frac{1}{6} \right| \leq \frac{\epsilon}{6} < \epsilon$$ The desired limit follows.