Calculate the limit : $\lim_{x \to 0}\frac{x-\sin{x}}{x^3}$ WITHOUT using L'Hopital's rule I was given a task to find  $$\lim_{x\to0}\frac{x-\sin{x}}{x^3}$$ at my school today. I thought it was an easy problem and started differentiating denominator and numerator to calculate the limit but the teacher then said we aren't allowed to use  L'Hopital's rule, but to "play around" with known limits and limit definition. I got stuck here since I can't really think of a way to do this, and according to my teacher, there are at least 4 ways. A subtle hint would be enough. Thanks.
 A: Note: This method is 'Taylor series inspired,' but I think I can say in good faith that it fits your requirements of not using l'Hospital's Rule or Taylor's Theorem.
We observe first that
$$\lim_{x\to 0^-} \frac{x-\sin x}{x^3} = \lim_{x\to0^+} \frac{(-x) - \sin(-x)}{(-x)^3} = \lim_{x\to0^+} \frac{x - \sin x}{x^3}$$
Thus, $$\lim_{x \to 0}\frac{x-\sin x}{x^3} = \lim_{x\to 0^+} \frac{x-\sin x}{x^3}$$
so we can focus on behavior of $\frac{x-\sin x}{x^3}$ for $x>0$ to evaluate the limit.
Set $f(x) = x - \sin x$.  We're going to try to bound $f$ with polynomials so we can use the Squeeze Theorem.  Since integration preserves inequalities, in the sense that if $f(t) \le g(t)$ for $0 \le t \le x$ then $\int_0^x f(t) \,dt \le \int_0^x g(t) \,dt$, we will find these polynomials by differentiating $f$ some number of times, finding polynomials that bound the derivative, then integrating. Since we want to find the limit of $\frac{f}{x^3}$, we'll differentiate $f$ three times, based on the reasoning that if $P(x) \le f^{(3)} \le Q(x)$ for polynomials $P$ and $Q$, then after integrating three times we will have an inequality involving polynomials with no terms of degree less than $3$. We calculate
$$f^{(3)}(x) = \cos x$$
For $x>0$, we have that
$$1-x \le \cos x \le 1$$
Let's integrate three times:
$$\int_0^x\int_0^y\int_0^z 1-w\, dw\,dz\,dy \le \int_0^x\int_0^y\int_0^z \cos w\, dw\,dz\,dy \le \int_0^x\int_0^y\int_0^z 1\, dw\,dz\,dy$$
Don't worry if you haven't seen triple integrals before: all we're doing is taking three integrals, one after the other.  This gives us
$$\frac{1}{6}x^3 - \frac{1}{24}x^4 \le x - \sin x \le \frac{1}{6}x^3$$
To finish, just have we divide by $x^3$ and take the limit as $x\to 0^+$.
A: $$L=\lim_{x\to0}\frac{x-\sin x}{x^3}\\
=\lim_{x\to0}\frac{2x-\sin2x}{8x^3}\\
4L-L = \lim_{x\to0}\frac{x-\frac12\sin2x-x+\sin x}{x^3}$$
which simplifies to the product of three expressions of the form $\frac{\sin y}{y}$
A: $\sin x = x - \dfrac{x^3}{3!} + o(x^5) \to \dfrac{x-\sin x}{x^3} = \dfrac{1}{6} + o(x^2)$. From this the limit is $\dfrac{1}{6}$.
A: I suppose you can always use Taylor expansion near $x=0$ (as long as it is known to you), that is:
$$
\sin x = x - \frac1{3!}x^3 + o(x^5)
$$
Hence we will have:
$$
\lim\limits_{x\to 0}\frac{x-\sin x }{x^3}=\frac16
$$
