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Let f(x) = {(x−1)^(1/(2−x)), x > 1, x ≠ 2; k, x = 2}. The value of k for which f is continuous at x = 2 is -

1. e^−2
2. e
3. e^−1
4. 1

Correct answer: e^−1

Solution

Let L = lim_{x->2} (x-1)^(1/(2-x)). Then ln L = lim (ln(x-1))/(2-x), a 0/0 form. By L'Hopital ln L = lim [1/(x-1)]/(-1) = -1. Hence k = L = e^-1.

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