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EXERCISE 8.1 · Q9

Q.Test the continuity of the following function at the point indicated against it: f(x)=x−1−(x−1)1/3x−2f(x) = \dfrac{\sqrt{x-1} - (x-1)^{1/3}}{x-2}, for x≠2x \ne 2, =15= \dfrac{1}{5}, for x=2x = 2, at x=2x = 2.

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f(x)=x−1−(x−1)1/3x−2f(x)=\dfrac{\sqrt{x-1}-(x-1)^{1/3}}{x-2} for x≠2x\ne2, and f(2)=15f(2)=\dfrac15.

Put u=x−1u=x-1, so x−2=u−1x-2=u-1; as x→2x\to2, u→1u\to1. Then f=u1/2−u1/3u−1f=\dfrac{u^{1/2}-u^{1/3}}{u-1}. Writing g(u)=u1/2−u1/3g(u)=u^{1/2}-u^{1/3}, note g(1)=1−1=0g(1)=1-1=0, so this is exactly the difference quotient g(u)−g(1)u−1\dfrac{g(u)-g(1)}{u-1}, whose limit as u→1u\to1 is the derivative g′(1)g'(1):

g′(u)=12u−1/2−13u−2/3,g′(1)=12−13=16.g'(u)=\tfrac12u^{-1/2}-\tfrac13u^{-2/3},\qquad g'(1)=\tfrac12-\tfrac13=\tfrac16.

So lim⁡x→2f(x)=16\displaystyle\lim_{x\to2} f(x)=\frac16. …

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