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7.5 I · Q63

Q.lim⁡x→asin⁡x−sin⁡ax5−a5\displaystyle\lim_{x\to a}\frac{\sin x-\sin a}{\sqrt[5]{x}-\sqrt[5]{a}}

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sin⁡x−sin⁡ax−a→cos⁡a\dfrac{\sin x-\sin a}{x-a}\to\cos a (a standard limit, provable by the substitution x=a+tx=a+t and the sum-to-product identity). Also x1/5−a1/5x−a→15a1/5−1=15a−4/5\dfrac{x^{1/5}-a^{1/5}}{x-a}\to\dfrac15a^{1/5-1}=\dfrac15a^{-4/5} by the standard power-difference theorem. Dividing the first by the second: cos⁡a(1/5)a−4/5=5a4/5cos⁡a\dfrac{\cos a}{(1/5)a^{-4/5}}=5a^{4/5}\cos a.

✓Final answer

5a4/5cos⁡a5a^{4/5}\cos a

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