Question

Determine the value of 'k' for which the following function is continuous at x = 3

`f(x) = {(((x + 3)^2 - 36)/(x - 3),  x != 3), (k,  x = 3):}`

Answer 1

LHL = `lim_(x rightarrow 3^-)f(x)`

= `lim_(x rightarrow 3^-)(((x + 3)^2 - 36)/(x - 3))`

= `lim_(h rightarrow 0)(((3 - h + 3)^2 - 36)/(3 - h - 3))`

Put x = 3 – h

= `lim_(h rightarrow 0)(((6 - h)^2 - 36)/-h)`

= `lim_(h rightarrow 0)((36 + h^2 - 12h - 36)/-h)`

= `lim_(h rightarrow 0)((h^2 - 12h)/-h)`

= `lim_(h rightarrow 0)(12 - h)`

= 12

RHL = `lim_(x rightarrow 3^+)f(x)`

= `lim_(x rightarrow 3^+) [((x + 3)^2 - 36)/(x - 3)]`

= `lim_(h rightarrow 0)[((3 + h + 3)^2 - 36)/(3 + h - 3)]`

Put x = 3 + h

= `lim_(h rightarrow 0)[((6 + h)^2 - 36)/h]`

= `lim_(h rightarrow 0)[(36 + h^2 + 12h - 36)/h]`

= `lim_(h rightarrow 0)(h + 12)`

= 12

For continuity,

LHL = RHL = f(3)

= 12 = 12 = k

Hence, k = 12