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Evaluate the following : limx→0{1x12[1-cos(x22)-cos(x44)+cos(x22)cos(x44)]} - Mathematics and Statistics

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Question

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)]}`

Sum
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Solution

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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Chapter 7: Limits - Miscellaneous Exercise 7.2 [Page 159]

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