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Question
Evaluate the following :
`lim_(x -> pi/6) [(cos x - sqrt(3) sinx)/(pi - 6x)]`
Solution
Put x = `pi/6 + "h"`. Then as `x -> pi/6, "h" -> 0`.
Also, π – 6x = `pi - 6(pi/6 + "h")` = π – π – 6h = – 6h
and `cos x - sqrt(3) sinx = cos (pi/6 + "h") - sqrt(3) sin (pi/6 + "h")`
= `(cos pi/6 cos "h" - sin pi/6 sin "h") - sqrt(3) (sin pi/6 cos "h" + cos pi/6 sin "h")`
= `((sqrt(3))/2 cos "h" - 1/2 sin "h") - sqrt(3) (1/2 cos "h" + sqrt(3)/2 sin "h")`
= `sqrt(3)/2 cos "h" - 1/2 sin "h" - sqrt(3)/2 cos "h" -3/2 sin "h"`
= – 2 sin h
∴ `lim_(x -> pi/6) (cos x - sqrt(3) sinx)/(pi - 6x)`
= `lim_("h" -> 0) (-2 sin "h")/(-6"h")`
= `1/3 lim_("h" -> 0) sin "h"/"h"`
= `1/3 xx 1`
= `1/3`
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