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Evaluate: limx→128x-32x-1-4x2+14x2-1 - Mathematics

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प्रश्न

Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`

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उत्तर

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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अध्याय 13: Limits and Derivatives - Exercise [पृष्ठ २४०]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 13 Limits and Derivatives
Exercise | Q 13 | पृष्ठ २४०

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