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Question
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`
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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`
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