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प्रश्न
Evaluate the following :
`lim_(x -> 0) {1/x^12 [1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)]}`
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उत्तर
Consider `1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)`
= `[1 - cos(x^2/2)] - cos(x^4/4) [1 - cos(x^2/2)]`
= `[1 - cos(x^2/2)] [1 - cos(x^4/4)]`
= `[1 - cos(x^2/2)][1 - cos(x^4/4)] xx (1 + cos(x^2/2))/(1 + cos(x^2/2)) xx (1 + cos(x^4/4))/(1 + cos(x^4/4))`
= `([1 - cos^2(x^2/2)][1 - cos^2 (x^4/4)])/([1 + cos(x^2/2)][1 + cos(x^4/4)]`
= `(sin^2(x^2/2) sin^2(x^4/4))/([1 + cos(x^2/2)][1 + cos(x^4/4)]`
∴ the required limit
= `lim_(x -> 0) (sin^2(x^2/2)sin^2(x^4/4))/(x^12[1 + cos(x^2/2)][1 + cos(x^4/4)]`
= `lim_(x -> 0) (sin^2(x^2/2) sin^2 (x^4/4))/(x^4/4 xx x^8/16 [1 + cos(x^2/2)][1 + cos(x^4/4)]) xx 1/64`
= `1/64([lim_(x -> 0) (sin(x^2/2))/((x^2/2))]^2 xx [lim_(x -> 0) (sin(x^4/4))/((x^4/4))]^2)/([lim_(x -> 0) {1 + cos(x^2/2)}] xx [lim_(x -> 0) {1 + cos(x^4/4)}]`
= `1/64 (1^2 xx 1^2)/((1 + cos 0)(1 + cos 0)) ...[(because x -> 0 therefore x^2/2 -> 0"," x^4/4 -> 0),("and" lim_(theta -> 0) sintheta/theta = 1)]`
= `1/64 xx 1/(2 xx 2)`
= `1/256`
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