Question

Evaluate the following :

`lim_(x -> pi/6) [(2 - sqrt(3)cosx - sinx)/(6x - pi)^2]`

Answer 1

Put x = `pi/6 + "h"`.

Then as `x -> pi/6, "h" -> 0`.

Also, 6x – π  = `6(pi/6 + "h") - pi` = π + 6h – π = 6h and 

`sqrt(3)cos x + sinx = sqrt(3)cos(pi/6 + "h") + sin(pi/6 + "h")`

= `sqrt(3)(cos  pi/6 cos "h" - sin  pi/6 sin"h") + (sin  pi/6 cos"h" + cos  pi/6 sin"h")`

= `sqrt(3)(sqrt(3)/2 cos"h" - 1/2 sin"h") + (1/2 cos"h" + sqrt(3)/2 sin"h")`

= `3/2 cos"h" - sqrt(3)/2 sin"h" + 1/2 cos"h" + sqrt(3)/2 sin"h"`

= 2 cos h

∴ `lim_(x -> pi/6) [(2 - sqrt(3)cosx - sinx)/(6x - pi)^2]`

= `lim_(x -> pi/6) (2 - (sqrt(3) cosx + sinx))/(6x - pi)^2`

= `lim_("h" -> 0) (2 - 2cos"h")/(6"h")^2`

= `lim_("h" -> 0) (2(1 - cos"h"))/(36"h"^2)`

= `lim_("h" -> 0) (1 - cos"h")/(18"h"^2) * (1 + cos"h")/(1 + cos"h")`

= `lim_("h" -> 0) (1 - cos^2"h")/(18"h"^2(1 + cos"h"))`

= `lim_("h" -> 0) (sin^2"h")/(18"h"^2(1 + cos"h"))`

= `1/18 lim_("h" -> 0) ((sin"h")/"h")^2 xx 1/(1 + cos "h")`

= `1/18(lim_("h" -> 0) (sin"h")/"h")^2 xx 1/(lim_("h" -> 0) (1 + cos"h")`

= `1/18(1)^2 xx 1/(1 + 1)`

= `1/36`