Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{1-\cos2\text{x}}{1-\cos\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^2\text{x}}{2\sin^2\frac{\text{x}}{2}}$
$=\frac{\lim\limits_{\text{x} \rightarrow0}(\sin\text{x})^2}{\lim\limits_{\text{x} \rightarrow0}\big(\sin\frac{\text{x}}{2}\big)^2}$
$=\frac{\lim\limits_{\text{x} \rightarrow0}\big(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\big)^2}{\lim\limits_{\text{x} \rightarrow0}\big(\sin\frac{\text{x}}{2}\big)^2}$
$=4\lim\limits_{\text{x} \rightarrow0}\cos^2\frac{\text{x}}{2}$
$=4\times1$
$=4$

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