Tìm giới hạn \(A=\lim\limits _{x \rightarrow 0} \frac{1+\sin m x-\cos m x}{1+\sin n x-\cos n x}\)

* Hướng dẫn giải

\(\begin{array}{c} \text { Ta có: } \dfrac{1+\sin m x-\cos m x}{1+\sin n x-\cos n x}=\dfrac{2 \sin ^{2} \frac{m x}{2}+2 \sin \frac{m x}{2} \cos \frac{m x}{2}}{2 \sin ^{2} \frac{n x}{2}+2 \sin \frac{n x}{2} \cos \frac{n x}{2}} \\ =\dfrac{m}{n}. \dfrac{\sin \dfrac{m x}{2}}{\dfrac{m x}{2}} \cdot \dfrac{\frac{n x}{2}}{\sin \dfrac{n x}{2}} \cdot \frac{\sin \dfrac{m x}{2}+\cos \dfrac{m x}{2}}{\sin \dfrac{n x}{2}+\cos \frac{n x}{2}} \\ A=\frac{m}{n} \lim \limits_{x \rightarrow 0} \dfrac{\sin \frac{m x}{2}}{\dfrac{m x}{2}} \cdot \lim\limits _{x \rightarrow 0} \frac{\frac{n x}{2}}{\sin \dfrac{n x}{2}} \cdot \lim\limits_{x \rightarrow 0} \dfrac{\sin \dfrac{m x}{2}+\cos \dfrac{m x}{2}}{\sin \dfrac{n x}{2}+\cos \dfrac{n x}{2}}=\dfrac{m}{n} \end{array}\)