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  • Evaluate the following limits limit x tends to 0 cos 2x - 1

17.   Evaluate the following limits  \lim_{x\rightarrow 0} \frac{\cos 2x -1}{\cos x -1}

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The limit:

  \lim_{x\rightarrow 0} \frac{\cos 2x -1}{\cos x -1}

The function takes the zero by zero form when the limit is put directly, so we simplify the function and then put the limit 

  \lim_{x\rightarrow 0} \frac{\cos 2x -1}{\cos x -1}

\lim_{x\rightarrow 0} \frac{-2(sin^2x)}{-2(sin^2(\frac{x}{2}))}

=\lim_{x\rightarrow 0}    \frac{\frac{(sin^2x)}{x^2}}{\frac{(sin^2(\frac{x}{2}))}{(\frac{x}{2})^2}}\times\frac{x^2}{(\frac{x}{2})^2}

=\frac{1^2}{1^2}\times 4

= 4  (Answer)

Posted by

Pankaj Sanodiya

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