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Question
Evaluate the following :
`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
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Solution
`lim_(x -> 0) (x(6^x - 3^x))/(cos (6x) - cos (4x))`
= `lim_(x -> 0) (x*3^x (2^x - 1))/(-2sin ((6x + 4x)/2) sin((6x - 4x)/2))`
= `1/(-2) lim_(x -> 0) (x*3^x (2^x - 1))/(sin 5x*sinx)`
= `1/(-2) lim_(x -> 0) ((x*3^x (2^x - 1))/x^2)/((sin 5x* sinx)/x^2) ...[("Divide Numerator and"),("Denominator by" x^2),(∵ x -> 0"," ∴ x ≠ 0 ∴ x^2 ≠ 0)]`
= `1/(-2) (lim_(x -> 0) [3^x ((2^x - 1))/x])/(lim_(x -> 0) ((sin5x)/x * sinx/x))`
= `1/(-2) (lim_(x -> 0)(3x) * lim_(x -> 0) ((2^x - 1)/x))/(lim_(x -> 0) (sinx/(5x) xx 5) * lim_(x -> 0) (sinx/x))`
= `1/(-2) xx (3^circ xx log 2)/(1 xx 5 xx 1) ...[(because x -> 0"," 5x -> 0),(lim_(x -> 0) ("a"^x - 1)/x = log "a"),(lim_(x -> 0) sinx/x = 1)]`
= `(-1)/10 log 2`
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