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प्रश्न
Evaluate the following:
`lim_(x -> pi/2) [(cos3x + 3cosx)/(2x - pi)^3]`
उत्तर
Put x = `pi/2 + "h"`.
Then as `x -> pi/2, "h" -> 0`.
Also, 2x – π = `2(pi/2 + "h") - pi` = π + 2h – π = 2h
∴ `lim_(x -> pi/2) [(cos3x + 3cosx)/(2x - pi)^3]`
= `lim_(x -> pi/2) (4cos^3x - 3cosx + 3cosx)/(2x - pi)^3`
= `lim_(x -> pi/2) (4cos^3x)/(2x - pi)^3`
= `lim_("h" -> 0) (4[cos (pi/2 + "h")]^3)/(2"h")^3`
= `lim_("h" -> 0) (4( - sin "h")^3)/(8"h"^3)`
= `-4/8 lim_("h" -> 0) (sin"h"/"h")^3`
= `-1/2 (lim_("h" -> 0) sin"h"/"h")^3`
= `-1/2(1)^3`
= `-1/2`
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