#### Question

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):}`

#### 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`

3 + 3 + 6 = k

k = 12

Is there an error in this question or solution?

Solution Determine the Value of 'K' for Which the Following Function is Continuous at X = 3 F(X)=(X+3)2−36/X−3 Concept: Concept of Continuity.