Discuss the continuity of the following function at the point indicated against them :

f(x) = `{:(=( sqrt(3) - tanx)/(pi - 3x)",", x ≠ pi/3),(= 3/4",", x = pi/3):}} "at" x = pi/3`

#### Solution

`"f"(pi/3) = 3/4` ...(Given) ...(1)

`lim_(x -> pi/3) "f"(x) = lim_(x -> pi/3) (sqrt(3) - tanx)/(pi - 3x)`

Put x = `pi/3 + "h"`.

Then as `x -> pi/3, "h" -> 0`.

π – 3x = `pi - 3(pi/3 + "h")` = – 3h and `tan pi/3 = sqrt(3)`

∴ tan x = `tan(pi/3 + "h")`

= `(tan (pi/3) + tan "h")/(1 - tan (pi/3)* tan "h")`

= `(sqrt(3) + tan "h")/(1 - sqrt(3)* tan "h")`

∴ `lim_(x -> pi/3) "f"(x) = lim_("h" -> 0) (sqrt(3) - (sqrt(3) + tan "h")/(1 - sqrt(3)* tan "h"))/(-3"h")`

= `lim_("h" -> 0) (sqrt(3) - 3 tan "h" - sqrt(3) - tan "h")/(-3"h"(1 - sqrt(3) * tan "h")`

= `lim_("h" -> 0) (-4 tan "h")/(-3"h"(1 - sqrt(3) tan "h")`

= `4/3 lim_("h" -> 0) (tan"h"/"h" 1/(1 - sqrt(3) * tan"h"))`

= `4/3(lim_("h" -> 0) tan"h"/"h") xx 1/(lim_("h" -> 0) (1 - sqrt(3) tan "h")`

= `4/3(1) xx 1/(1 - sqrt(3) xx 0)` ...[h → 0, tan h →** **0]

= `4/3` ...(2)

From (1) and (2),

`lim_(x -> pi/3) "f"(x) ≠ "f"(pi/3)`

∴ f is discontinuous at x = `pi/3`