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Question
Evaluate the following :
`lim_(x -> pi/4) [(sqrt(2) - cosx - sinx)/(4x - pi)^2]`
Solution
`lim_(x -> pi/4) [(sqrt(2) - cosx - sinx)/(4x - pi)^2]`
= `lim_(x -> pi/4) [(sqrt(2) - (cos x + sinx))/[4(x - pi/4)]^2]`
Put `x - pi/4` = h
∴ x = `pi/4 + "h"`
As `x -> pi/4, "h" -> 0`
Also,
cos x + sin x = `cosx + cos(pi/2 - x)`
= `cos(pi/4 + "h") + cos[pi/2 - (pi/4 + "h")]`
= `cos(pi/4 + "h") + cos(pi/4 - "h")`
= `2cos pi/4.cos"h"` ...(By defactorisation)
= `2(1/sqrt(2))cos"h"`
= `sqrt(2)cos"h"`
∴ Required limit
= `lim_("h" -> 0) (sqrt(2) - sqrt(2)cos"h")/(4"h")^2`
= `lim_("h" -> 0) (sqrt(2) (1 - cos"h"))/(16"h"^2)`
= `sqrt(2)/16 lim_("h" -> 0) ((1 - cos"h")/"h"^2 xx (1 + cos"h")/(1 + cos"h"))`
= `1/(8sqrt(2)) lim_("h" -> 0) (sin^2"h")/("h"^2 (1 + cos"h"))`
= `1/(8sqrt(2)) lim_("h" -> 0) ((sin"h")/"h")^2 xx 1/(1 + cos"h")`
= `1/(8sqrt(2)) lim_("h" -> 0) ((sin"h")/"h")^2 xx lim_("h" -> 0) 1/(1 + cos"h")`
= `1/(8sqrt(2)) xx (1)^2 xx 1/(1 +1)`
= `1/(8sqrt(2)) xx 1/2`
= `1/(16sqrt(2))`
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