Sum
Determine the value of 'k' for which the follwoing function is continuous at x = 3
`f(x) = {(((x+3)^2-36)/(x-3), x != 3), (k, x =3):}`
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Solution
Given f(x) is continuous at x = 3
`:. lim_(x->3) f(x)= k`
`lim_(x-> 3) ((x+3)^2 - 36)/(x-3) = k`
`lim_(x->3) ((x+3)^2 - 6^2)/(x-3) = k`
`lim_(x->3) ((x+3+6)(x+3-6))/(x-3) = k`
`lim_(x->3) ((x - 3)(x + 9))/(x-3)`
`lim_(x->3) (x + 9) = 3 + 9`
k = 12
Concept: Concept of Continuity
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