Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-1}$

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-1}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{1-2\sin^2\big(\frac{\text{ax}}{2}\big)-1+2\sin^2\big(\frac{\text{bx}}{2}\big)}{1-2\sin^2\big(\frac{\text{cx}}{2}\big)-1}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{-2\sin^2\big(\frac{\text{ax}}{2}\big)+2\sin^2\big(\frac{\text{bx}}{2}\big)}{-2\sin^2\big(\frac{\text{cx}}{2}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{-\sin^2\big(\frac{\text{ax}}{2\text{ax}}\big)4\text{a}^2\text{x}^2+\sin\big(\frac{\text{bx}}{2}\big)4\text{b}^2\text{x}^2}{-\sin^2\big(\frac{\text{cx}}{2}\big)4\text{c}^2\text{x}^2}$ $=\frac{-\text{a}^2+\text{b}^2}{-\text{c}^2}$ $=\frac{\text{a}^2-\text{b}^2}{\text{c}^2}$

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