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Evaluate: limx→02sinx-sin2xx3

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प्रश्न

Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`

योग
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उत्तर

Given that `lim_(x -> 0) (2 sin x - sin 2x)/x^3`

= `lim_(x -> 0) (2 sin x - 2 sin x cos x)/x^3`

= `lim_(x -> 0) (2 sin x(1 - cosx))/x^3`

= `lim_(x -> 0) (2sinx)x (( - cosx)/x)`

= `lim_(x -> ) ((sinx)/x)((sin^2 x/2)/x^2)`

= `lim_(x -> 0) 2((sinx)/x)(2 (sin^2  x/2)/(x^2/4) xx 1/4)`

= `lim_(x -> 0) 2((sin x)/x) 2[((sin  x/2)/(x/2))^2] * 1/4`

= `lim_(x -> 0) 4/4 ((sin x)x)`

= `lim_(x/2 -> 0) ((sin x/2)/(x/2))^2`

= `1 * 1 * (1)^2`

= 1   .....`[because  lim_(x -> 0)  sinx/x = 1]`

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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 18 | पृष्ठ २४०

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