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Question
If f (x) = `(1 - "sin x")/(pi - "2x")^2` , for x ≠ `pi/2` is continuous at x = `pi/4` , then find `"f"(pi/2) .`
Solution
∵ f (x) is continuous at x = `pi/2`.
We have
`"f"(pi/2) = lim_(x -> pi/2) "f(x)"`
`= lim_(x -> pi/2) (1 - "sin x")/(pi - 2"x")^2`
Put x = `pi/2 + "h"`
so that as `"x" -> pi/2` , h → 0
`= lim_(h -> 0) (1 - "sin" (pi/2 + "h"))/[[pi - 2 (pi/2 + "h")]]`
`= lim_(h -> 0) (1 - "cos h")/(4"h"^2)`
`= lim_(h -> 0) (2 "sin"^2 ("h"/2))/(4"h"^2)`
`= lim_(h -> 0) ("sin"^2 ("h"/2))/(("h"^2/4) xx 4) xx 1/2`
`= 1/2 xx 1 xx 1/4`
`= 1/8`
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