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प्रश्न
Evaluate the following :
`lim_(x -> pi) [(sqrt(5 + cosx) - 2)/(pi - x)^2]`
उत्तर
`lim_(x -> pi) [(sqrt(5 + cosx) - 2)/(pi - x)^2]`
Put π – x = h
∴ x = π – h
As x → π, h → 0
∴ `lim_(x -> pi) [(sqrt(5 + cosx) - 2)/(pi - x)^2]`
= `lim_("h" -> 0) [(sqrt(5 + cos (pi - "h")) - 2)/("h")^2]`
= `lim_("h" -> 0) [(sqrt(5 - cos "h") - 2)/"h"^2]`
= `lim_("h" -> 0) [(sqrt(5 - cos "h") - 2)/"h"^2 xx (sqrt(5 - cos"h") + 2)/(sqrt(5 - cos"h") + 2)]`
= `lim_("h" -> 0) (5 - cos"h" - 4)/("h"^2(sqrt(5 - cos"h") + 2))`
= `lim_("h" -> 0) (1 - cos"h")/("h"^2(sqrt(5 - cos"h") + 2))`
= `lim_("h" -> 0) (1 - cos"h")/("h"^2(sqrt(5 - cos"h") + 2)) xx (1 + cos"h")/(1 + cos"h")`
= `lim_("h" -> 0) (sin^2"h")/("h"^2(sqrt(5 - cos"h") + 2)(1 + cos"h"))`
= `lim_("h" -> 0) (sin"h"/"h")^2 xx 1/((sqrt(5 - cos"h") + 2)(1 + cos"h"))`
= `lim_("h" -> 0) (sin"h"/"h")^2 xx lim_("h" -> 0) 1/((sqrt(5 - cos"h") + 2)(1 + cos"h"))`
= `(1)^2 xx 1/((sqrt(5 - 1) + 2)(1 + 1))`
= `1/((sqrt(4) + 2)(2)`
= `1/((2 + 2)(2))`
= `1/8`
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