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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Evaluate the following limits \[\] $ \lim\limits_{ x \to 0} \large\frac{\cos\: 2x-1}{\cos\: x - 1}$

$\begin{array}{1 1}4 \\ 1 \\ 0 \\ 2 \end{array} $

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At $ x = 0 $ the value of the given function is $ \large\frac{0}{0}.$
We know that: $\quad \cos \: x = 1-2 sin^2 \large\frac{x}{2} $
$\Rightarrow$ $ \lim\limits_{ x \to 0} \large\frac{\cos\: 2x-1}{\cos\: x - 1}$$ = \lim\limits_{x \to 0} \large\frac{1-2 \sin^2x-1}{1-2 \sin^2 \large\frac{x}{2} -\normalsize1}$ $ = \lim\limits_{x \to 0} \large\frac{ \sin^2x}{ \sin^2 \large\frac{x}{2} }$
$\large\frac{\sin^2x}{\sin^2 \large\frac{x}{2}} $ $ =\Large\frac{ \Large\frac{\sin^2x}{x^2} \normalsize \times x^2}{ \Large\frac{\sin^2 \large\frac{x}{2}}{ \large\frac{x}{2} ^2} \normalsize \times \Large\frac{x^2}{4}}$$=$$ 4 \times \Large\frac { \Large\frac{\sin^2 x}{x^2} } { \Large\frac {\sin^2 \large\frac{x}{2} }{\Large\frac{x^2}{4}} }$
$\Rightarrow$ $ \lim\limits_{ x \to 0} \large\frac{\cos\: 2x-1}{\cos\: x - 1}$$ = 4 \large\frac{ \lim\limits_{x \to 0} \bigg( \Large\frac{sin^2x}{x^2} \bigg)}{ \lim\limits_{x \to 0} \bigg( \Large\frac{sin^2 \Large\frac{x}{2}}{ \Large\frac{x}{2}^4}\bigg)}$
As $x \rightarrow 0 \Rightarrow \large\frac{x}{2} $$\rightarrow 0$
$\Rightarrow$ $ \lim\limits_{ x \to 0} \large\frac{\cos\: 2x-1}{\cos\: x - 1}$$ =4 \large\frac{\bigg( \lim\limits_{x \to 0}\Large\frac{sin\: x}{x} \bigg)^2}{ \lim\limits_{ \large\frac{x}{2} \to 0} \bigg( \Large\frac{sin \Large\frac{x}{2}}{ \Large\frac{x}{2} } \bigg)^2}$
$\qquad = 4, \qquad $ Using $\lim\limits_{y \to 0} \large\frac{ \sin\: y}{y}$$ = 1 $
answered Apr 4, 2014 by thanvigandhi_1
edited May 16, 2014 by balaji.thirumalai
 

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