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Question
Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`
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Solution
Given that `lim_(x -> 1/2) ((8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1))`
= `lim_(x -> 1/2) [((8x - 3)(2x + 1) - (4x^2 + 1))/((4x^2 - 1))]`
= `lim_(x -> 1/2) [(16x^2 - 6x + 8x - 3 - 4x^2 - 1)/(4x^2 - 1)]`
= `lim_(x -> 1/2) [(12x^2 + 2x - 4)/(4x^2 - 1)]`
= `lim_(x -> 1/2) (2(6x^2 + x - 2))/(4x^2 - 1)`
= `lim_(x -> 1/2) (2[6x^2 + 4x - 3x - 2])/((2x + 1)(2x - 1))`
= `lim_(x -> 1/2) (2[2x(3x + 2) - 1(3x + 2)])/((2x + 1)(2x - 1))`
= `lim_(x -> 1/2) (2(3x + 2)(2x - 1))/((2x + 1)(2x - 1))`
= `lim_(x -> 1/2) (2(3x + 2))/((2x + 1))`
Taking limit, we have
= `(2(3 xx 1/2 + 2))/(2 xx 1/2 + 1)`
= `(2(7/2))/2`
= `7/2`
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