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Question
Select the correct answer from the given alternatives.
`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =
Options
`3/2`
`1/2`
`-1/2`
`1/4`
Solution
`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3] = underline (-1/2)`
Explanation:
`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]`
= `lim_(x -> pi/2)(3cos x + 4cos^3x - 3cos x)/(2x - pi)^3`
= `lim_(x -> pi/2) (4cos^3x) /(8(x - pi/2)^3`
Put `x = pi/2 + h,`
`x - pi/2 = h`
As `x -> pi/2, h -> 0`
= `lim_(x -> pi/2) (4 cos^3x) /(8(x - pi/2)^3`
= `lim_(h-> 0) (4 cos^3(pi/2 + h))/(8h^3)`
= `lim_(h -> 0) (4(-sin h)^3)/(8h^3)` ...`[∵ cos(pi/2 + θ) = -sinθ]`
= `-1/2(lim_(h->0)(sin h)/h)^3`
= `-1/2`
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