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Exercise: Limits of Trigonometric Fun... · Q19

Q.Evaluate lim⁡x→π/21−sin⁡xcos⁡2x\lim_{x\to\pi/2}\dfrac{1-\sin x}{\cos^2x}.

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At x=π/2x=\pi/2: 1−sin⁡x→01-\sin x\to0 and cos⁡2x→0\cos^2x\to0, a 0/00/0 form. Using cos⁡2x=1−sin⁡2x=(1−sin⁡x)(1+sin⁡x)\cos^2x=1-\sin^2x=(1-\sin x)(1+\sin x),

1−sin⁡xcos⁡2x=1−sin⁡x(1−sin⁡x)(1+sin⁡x)=11+sin⁡x(sin⁡x≠1).\frac{1-\sin x}{\cos^2x} = \frac{1-\sin x}{(1-\sin x)(1+\sin x)} = \frac{1}{1+\sin x} \quad (\sin x\neq1). …

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