Example Question - evaluating limit

Evaluating the Limit of a Trigonometric Function

\[ \lim_{{x \to 0}} \frac{{\sin(x) - x}}{{x^3}} = \lim_{{x \to 0}} \frac{{\sin(x) - x}}{{x^3}} \cdot \frac{{\sin(x) + x}}{{\sin(x) + x}} = \lim_{{x \to 0}} \frac{{\sin^2(x) - x^2}}{{x^3(\sin(x) + x)}} \] \[ = \lim_{{x \to 0}} \frac{{\sin^2(x) - x^2}}{{x^3}} \cdot \lim_{{x \to 0}} \frac{1}{{\sin(x) + x}} \] \[ = \lim_{{x \to 0}} \frac{{(\sin(x) + x)(\sin(x) - x)}}{{x^3}} \cdot \lim_{{x \to 0}} \frac{1}{{\sin(x) + x}} \] Using the identity $(a+b)(a-b) = a^2 - b^2$ and L'Hôpital's Rule, \[ = \lim_{{x \to 0}} \frac{{\sin(x) - x}}{{x^2}} \cdot \lim_{{x \to 0}} \frac{1}{{\sin(x) + x}} \] Applying L'Hôpital's Rule for the first limit, \[ = \lim_{{x \to 0}} \frac{{\cos(x) - 1}}{{2x}} \] \[ = \frac{1}{2} \cdot \lim_{{x \to 0}} \frac{{\cos(x) - 1}}{{x}} \] Applying L'Hôpital's Rule once more, \[ = \frac{1}{2} \cdot \lim_{{x \to 0}} \frac{{-\sin(x)}}{1} \] \[ = \frac{1}{2} \cdot \lim_{{x \to 0}} -\sin(x) \] \[ = \frac{1}{2} \cdot 0 \] \[ = 0 \]