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प्रश्न
Evaluate the following :
`lim_(x -> 1) [(1 - x^2)/(sinpix)]`
उत्तर
`lim_(x -> 1) (1 - x^2)/(sinpix) = (1 - x 1 + x)/(sinpix)`
Put 1 – x = h,
∴ x = 1 – h
As x → 1, h → 0
∴ `lim_(x -> 1) (1 - x 1 + x)/(sinpix)`
= `lim_("h" -> 0) (("h")(1 + 1 - "h"))/(sin pi (1 - "h"))`
= `lim_("h" -> 0) (("h")(2 - "h"))/(sin (pi - pi"h"))`
= `lim_("h" -> 0) (("h")(2 - "h"))/(sin pi"h")` ...[∵ sin (π – θ) = sin θ]
= `lim_("h" -> 0) ((2 - "h"))/(((sin pi"h")/(pi"h"))*pi)`
= `1/pi * (lim_("h" -> 0) (2 - "h"))/(lim_("h" -> 0) ((sin pi"h")/(pi"h"))`
= `1/pi* ((2 - 0))/1 ...[because "h" -> 0"," pi"h" -> 0 "and" lim_(theta -> 0) (sintheta)/theta = 1]`
= `2/pi`
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