Question-9912

Question Number 9912 by ridwan balatif last updated on 15/Jan/17

Answered by mrW1 last updated on 15/Jan/17

\lim_{x \to π} \frac{1 + \cos x}{{(x - π)}^{2}}

t = x - π

x = t + π

1 + \cos x = 1 + \cos (t + π) = 1 - \cos t

= 1 - (1 - {2 \sin}^{2} \frac{t}{2}) = {2 \sin}^{2} (\frac{t}{2})

\lim_{x \to π} \frac{1 + \cos x}{{(x - π)}^{2}} = \lim_{t \to 0} \frac{{2 \sin}^{2} (\frac{t}{2})}{t^{2}} = \frac{1}{2} \times \lim_{t \to 0} {[\frac{\sin (\frac{t}{2})}{(\frac{t}{2})}]}^{2} = \frac{1}{2} \times 1^{2} = \frac{1}{2}

\Rightarrow A n s w e r (E)

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