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Question
If f(x) = `(sqrt(2 + sin x) - sqrt(3))/(cos^2x), "for" x ≠ pi/2`, is continuous at x = `pi/2` then find `"f"(pi/2)`
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Solution
f is given to be continuous at x = `pi/2`
∴ by defination,
`"f"(pi/2) = lim_(x -> pi/2) "f"(x)`
= `lim_(x -> pi/2) (sqrt(2 + sinx) - sqrt(3))/(cos^2x)`
= `lim_(x -> pi/2) (sqrt(2 + sinx) - sqrt(3))/(cos^2x) xx (sqrt(2 + sinx) + sqrt(3))/(sqrt(2 + sinx) + sqrt(3))`
= `lim_(x -> pi/2) ((2 + sin x) - 3)/((1 - sin^2x)(sqrt(2 + sin x) + sqrt(3))`
= `lim_(x -> pi/2) (-(1 - sin x))/((1 - sin x)(1 + sin x)(sqrt(2 + sin x) + sqrt(3))`
= `lim_(x -> pi/2) (-1)/((1 + sin x)[sqrt(2 + sinx) + sqrt(3)]) ...[because x -> pi/2, x ≠ pi/2 therefore sin x ≠ sin pi/2 = 1 therefore 1 - sin x ≠ 0]`
= `(lim_(x -> pi/2) ( - 1))/([lim_(x -> pi/2) (1 + sin x)] xx [lim_(x -> pi/2) (sqrt(2 + sin x) + sqrt(3)]`
= `(-1)/((1 + sin pi/2) (sqrt(2 + sin pi/2) + sqrt(3))`
= `(-1)/((1 + 1)(sqrt(2 + 1) + sqrt(3))`
∴ `"f"(pi/2) = (-1)/(4sqrt(3))`
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