Tính giới hạn \(\lim \limits_{x \rightarrow 0} \frac{\cos a x-\cos b x \cos c x}{\sin ^{2} x}\)
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Lời giải:
Báo saiTa có
\(\frac{\cos a x-\cos b x \cos c x}{\sin ^{2} x}=[\cos a x-\cos b x+\cos b x(1-\cos c x)] \cdot \frac{1}{\sin ^{2} x}\)
\(\begin{aligned} &=\left[-2 \sin \left(\frac{a x+b x}{2}\right) \sin \left(\frac{a x-b x}{2}\right)+2 \cos b x \sin ^{2} \frac{c x}{2}\right] \cdot \frac{1}{\sin ^{2} x} \\ &=\left[-2 \cdot \frac{\sin \left(\frac{a x+b x}{2}\right)}{\frac{a x+b x}{2}} \cdot \frac{\sin \left(\frac{a x-b x}{2}\right)}{\frac{a x-b x}{2}} \cdot \frac{a x+b x}{2} \cdot \frac{a x-b x}{2}+2 \cos b x\left(\frac{\sin \frac{c x}{2}}{\frac{c x}{2}}\right)^{2} \cdot \frac{c^{2} x^{2}}{4}\right] . \left(\frac{x}{\sin x}\right)^{2} \cdot \frac{1}{x^{2}} \end{aligned}\)
\(=\left[\frac{b^{2}-a^{2}}{2} \cdot \frac{\sin \left(\frac{a x+b x}{2}\right)}{\frac{a x+b x}{2}} \cdot \frac{\sin \left(\frac{a x-b x}{2}\right)}{\frac{a x-b x}{2}}+\cos b x\left(\frac{\sin \frac{c x}{2}}{\frac{c x}{2}}\right)^{2} \cdot \frac{c^{2}}{2}\right] \cdot\left(\frac{x}{\sin x}\right)^{2} .\)
\(\text { Do đó: } \lim _{x \rightarrow 0} \frac{\cos a x-\cos b x \cos c x}{\sin ^{2} x}=\frac{b^{2}+c^{2}-a^{2}}{2} \text { . }\)
