Calculate the limit : $\lim_{x \to 0}\frac{x-\sin{x}}{x^3}$ WITHOUT using L'Hopital's rule [duplicate]
$$L=\lim_{x\to0}\frac{x-\sin x}{x^3}\\
=\lim_{x\to0}\frac{2x-\sin2x}{8x^3}\\
4L-L = \lim_{x\to0}\frac{x-\frac12\sin2x-x+\sin x}{x^3}$$
which simplifies to the product of three expressions of the form $\frac{\sin y}{y}$
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 $x\to 0^+$.