Question - Evaluating the Limit of a Trigonometric Function
Solution:
$$\lim_{{x \to 0}} \frac{{\sin(x) - x}}{{x^3}} = \lim_{{x \to 0}} \frac{{\sin(x) - x}}{{x^3}} \cdot \frac{{\sin(x) + x}}{{\sin(x) + x}} = \lim_{{x \to 0}} \frac{{\sin^2(x) - x^2}}{{x^3(\sin(x) + x)}}$$$$= \lim_{{x \to 0}} \frac{{\sin^2(x) - x^2}}{{x^3}} \cdot \lim_{{x \to 0}} \frac{1}{{\sin(x) + x}}$$$$= \lim_{{x \to 0}} \frac{{(\sin(x) + x)(\sin(x) - x)}}{{x^3}} \cdot \lim_{{x \to 0}} \frac{1}{{\sin(x) + x}}$$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,$$= \frac{1}{2} \cdot \lim_{{x \to 0}} \frac{{-\sin(x)}}{1}$$$$= \frac{1}{2} \cdot \lim_{{x \to 0}} -\sin(x)$$$$= \frac{1}{2} \cdot 0$$$$= 0$$
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