calculus - Algebra of limits vs. L'Hospital

We have to evaluate the following limit:

$$\displaystyle\lim_{x\to 0} \dfrac{x - \sin(x)}{x^3}$$

When I evaluate it using L'Hospital rule I get $\dfrac16$ but when I simplify the limit using simple algebraic rules of limit I get $0$.

We can use the this rule:
$\displaystyle\lim [f(x) \pm g(x)] = \lim[f(x)] \pm \lim[g(x)]$ to simplify the given limit.

$$\lim_{x\to 0} \dfrac{x - \sin(x)}{x^3} = \lim_{x\to 0}\dfrac{x}{x^3} - \lim_{x\to 0}\dfrac{\sin(x)}{x^3}$$

Also, $\lim[f(x).g(x)] = \lim[f(x)].\lim[g(x)]$.

$$\therefore \lim_{x\to 0}\dfrac{x}{x^3} - \lim_{x\to 0}\dfrac{\sin(x)}{x^3} = \lim_{x \to 0}\dfrac{1}{x^2} - \left(\lim_{x\to 0} \dfrac{\sin(x)}{x} \right)\left(\lim_{x\to 0} \dfrac{1}{x^2}\right)$$

Now, since, $\lim_{x\to 0} \dfrac{\sin(x)}{x} = 1$,

$$\lim_{x \to 0}\dfrac{1}{x^2} - \left(\lim_{x\to 0} \dfrac{\sin(x)}{x} \right)\left(\lim_{x\to 0} \dfrac{1}{x^2}\right) = \lim_{x \to 0}\dfrac{1}{x^2} - \lim_{x\to 0} \dfrac{1}{x^2} = \lim_{x \to 0}\left(\dfrac{1}{x^2} - \dfrac{1}{x^2}\right) = 0$$

Why am I getting $0$ when I use algebra of limits.

Answer

Rules like

$$\lim[f(x) \pm g(x)]=\lim[f(x)] \pm \lim[g(x)]$$

are only true if all the limits involved converge to finite numbers. Almost all the limits involved in your argument turn out to approach infinity.