If 0 leq x leq pi / 2,then lim _{x rightarrow | vedclass

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(D) Let $f(x) = \frac{|2 \cos x - 1|}{2 \cos x - 1}$ for $0 \leq x \leq \pi/2$. We know that $2 \cos x - 1 > 0$ when $\cos x > 1/2$,i.e.,$0 \leq x < \

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$=1$,when $0 \leq a < \pi/3$

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(D) Let $f(x) = \frac{|2 \cos x - 1|}{2 \cos x - 1}$ for $0 \leq x \leq \pi/2$. We know that $2 \cos x - 1 > 0$ when $\cos x > 1/2$,i.e.,$0 \leq x Thus,$f(x) = 1$ for $0 \leq x For the limit to exist at $x=a$,the left-hand limit and right-hand limit must be equal. For $0 \leq a For $\pi/3 At $a = \pi/3$,the left-hand limit is $1$ and the right-hand limit is $-1$,so the limit does not exist. Therefore,the correct statement is that the limit is $1$ when $0 \leq a < \pi/3$.

If $0 \leq x \leq \pi / 2$,then $\lim _{x \rightarrow a} \frac{|2 \cos x-1|}{2 \cos x-1}$

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