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Question
Evaluate the following :
`lim_(x -> pi/2) [("cosec"x - 1)/(pi/2 - x)^2]`
Solution
Put `pi/2 - x` = h,
∴ x = `pi/2 - "h"`
As `x -> pi/2"`, `"h" -> 0`
∴ `lim_(x -> pi/2) ("cosec" x - 1)/(pi/2 - x)^2 = lim_("h" -> 0) ("cosec"(pi/2 - "h") - 1)/"h"^2`
= `lim_("h" -> 0) (sec"h" - 1)/"h"^2 ...[because "cosec" (pi/2 - theta) = sec theta]`
= `lim_("h" -> 0) (1/cos"h" - 1)/"h"^2`
= `lim_("h" -> 0) (1 - cos"h")/("h"^2* cos"h")`
= `lim_("h" -> 0) (2sin^2 "h"/2)/"h"^2 * 1/cos"h"`
= `2 lim_("h" -> 0) ((sin "h"/2)/("h"/2))^2 * 1/4 * 1/cos"h"`
= `1/2 lim_("h" -> 0) ((sin "h"/2)/("h"/2))^2 lim_("h" -> 0) 1/(cos "h"`
= `1/2(1)^2 xx 1/cos0 ...[because "h" -> 0, "h"/2 -> 0 "and" lim_(theta -> 0) sintheta/theta = 1]`
= `1/2 xx 1`
= `1/2`
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