Evaluate the following : limx→π[5+cosx-2(π-x)2] - Mathematics and Statistics

Sum

Evaluate the following :

`lim_(x -> pi) [(sqrt(5 + cosx) - 2)/(pi - x)^2]`

`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`

Concept: Substitution Method

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Chapter 7 Limits
Exercise 7.5 | Q I. (3) | Page 150