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.
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