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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):}`
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Solution
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
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