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Question:

Let f(x) = \begin{cases} 12 - x, & x > 1, x \neq 2 \ k, & x = 2 \end{cases} The value of k for which f is continuous at x = 2 is

e

e^x

e^2

1

Solution:

If f(x) is continuous at x = 2, then limx→2(12 - x) = k
Above is 1∞ form, therefore k = el where l = limx→2(x - 2) × (12 - x) = limx→2 x - 2 = 0 ⇒ k = e0 = 1