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प्रश्न
If f(x) = `(cos^2 x - sin^2 x - 1)/(sqrt(3x^2 + 1) - 1)` for x ≠ 0, is continuous at x = 0 then find f(0)
उत्तर
f(x) is continuous at x = 0 ...(Given)
∴ f(0) = `lim_(x -> 0) "f"(x)`
= `lim_(x -> 0) (cos^2 x - sin^2x - 1)/(sqrt(3x^2 + 1) - 1)`
= `lim_(x -> 0) (cos2x - 1)/(sqrt(3x^2 + 1) - 1) xx (sqrt(3x^2 + 1) + 1)/(sqrt(3x^2 + 1) + 1)`
= `lim_(x -> 0) (-(1 - cos 2x) (sqrt(3x^2 + 1) + 1))/((3x^2 + 1) - 1)`
= `lim_(x -> 0) (-2sin^2x * (sqrt(3x^2 + 1) + 1))/(3x^2)`
= `(-2)/(3) lim_(x -> 0) (sin^2x)/(x^2) (sqrt(3x^2 + 1) + 1)`
= `(-2)/(3) lim_(x -> 0) (sinx/x)^2 xx lim_(x -> 0) (sqrt(3x^2 + 1) + 1)`
= `(-2)/(3) (1)^2 xx (sqrt(3(0) + 1) + 1)`
= `(-2)/(3) xx (1 + 1)`
∴ f(0) = `(-4)/(3)`
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