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Miscellaneous 7 II · Q124

Q.lim⁡x→a[sin⁡x−sin⁡ax−a]\displaystyle\lim_{x\to a}\left[\frac{\sin x-\sin a}{x-a}\right]

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Using sin⁡x−sin⁡a=2cos⁡ ⁣(x+a2)sin⁡ ⁣(x−a2)\sin x-\sin a=2\cos\!\left(\dfrac{x+a}2\right)\sin\!\left(\dfrac{x-a}2\right), and dividing by x−ax-a: $\dfrac{2\cos!\left(\tfrac{x+a}2\right)\sin!\left(\tfrac{x-a}2\right)}{x-a}=\cos!\left(\tfrac{x+a}2\right)\cdot\dfrac{ …

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