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