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Question
Discuss the continuity of the function f at x = 0, where
f(x) = `(5^x + 5^-x - 2)/(cos2x - cos6x),` for x ≠ 0
= `1/8(log 5)^2,` for x = 0
Solution
`lim_(x → 0) f(x) = lim_(x → 0)[(5^x + 5^-x - 2)/(cos2x - cos6x)]`
= `lim_(x → 0)[(5^x + 1/5^x - 2)/(2sin4x . sin2x)]`
= `lim_(x → 0)[(5^(2x) + 1 - 2(5^x))/(5^x . 2.sin 4x .sin2x)]`
= `1/2lim_(x → 0)[(5^x - 1)^2/(5^x.sin4x.sin2x)]`
= `1/2lim_(x → 0)[((5^x - 1)/x)^2/(5^x((sin4x)/(4x)).4((sin2x)/(2x)).2)]`
= `1/16 . (lim_(x → 0)((5^x - 1)/x)^2)/(lim_(x → 0)(5^x).lim_(x → 0)((sin4x)/(4x)).lim_(x → 0)((sin2x)/(2x))`
= `1/16 . (log 5)^2/(5^0(1)(1))`
∴ `lim_(x → 0) f(x) = 1/16 (log 5)^2`
= `1/16 xx 2 log 5`
= `1/8 log 5`
Given `f(0) = 1/8 (log5)^2`
∴ `lim_(x → 0) f(x) ≠ f(0)`
∴ f is discontinuous at x = 0.
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