Evaluate the following limits in Exercise: _ x 0 2x-1 -1 - Vidyadip

Question

Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$

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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$At x = 0, the value of the given function takes the form${\frac{0}{0}}$.
$\text{Now},$$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}=\lim\limits_{\text{x}\rightarrow0}\frac{1-2\sin^2\text{x}-1}{1-2\sin^2\frac{\text{x}}{2}-1} \Big[\cos\text{x}=1-2\text{sin}^2\frac{\text{x}}{2}\Big] $
$=\lim\limits_{\text{x}\rightarrow0}\frac{1-\sin^2\text{x}}{1-\sin^2\frac{\text{x}}{2}}=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\frac{\sin^2\text{x}}{\text{x}^2}\big)\times{\text{x}^2}}{\Bigg(\frac{\sin^2\frac{\text{x}}{2}}{\big(\frac{\text{x}}{2}\big)^2}\Bigg)\times\frac{\text{x}^2}{4}}$
$=4\frac{\lim\limits_{\text{x}\rightarrow0}\big(\frac{\sin^2\text{x}}{\text{x}^2}\big)}{\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{\sin^2\frac{\text{x}^2}{2}}{\big(\frac{\text{x}}{2}\big)^2}\Bigg)} $
$=4\frac{\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\bigg)^2}{\Bigg(\lim\limits_{\frac{\text{x}}{2}\rightarrow0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\Bigg)^2} \big[\text{x}\rightarrow0\Rightarrow\frac{\text{x}}{2}\rightarrow0\big] $ $=4\frac{1^2}{1^2} \big[\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{\text{y}}=1\big]$$=4$

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