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$\underset{x\to 0}{\lim} \dfrac{(1-\cos x)}{2}$ is equal to

  1. $0$
  2. $1$
  3. $1/3$
  4. $1/2$
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2 Answers

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Correct answer is A: 0

Since it is not 0/0 condition so we can't apply L'Hospital Rule and just putting x=0 will give 0 and also when expanding cos x=cos^2 (x/2) - sin^2 (x/2) and then putting x=0 will still give 0(zero) as answer.
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$\large \lim_{x->0}\frac{\left ( 1-cosx \right )}{2}$

=>$\large \lim_{x->0}\frac{\left ( 1-cosx \right )}{x}*\frac{x}{2}$

=$\large \lim_{x->0}\frac{\left ( 1-cosx \right )}{x}\lim_{x->0}*\frac{x}{2}$   [$\large \lim_{x->0}\frac{\left ( 1-cosx \right )}{x}=0$]

=$0*0$

=$0$
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