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A function is defined as: f(x) = (1 - cos(4x)) / x² for x < 0; f(0) = a; f(x) = sqrt(x) / (sqrt(16 + sqrt(x)) - 4) for x > 0. For what value of a is f(x) continuous at x = 0?
- f(x) is continuous when a = 0
- f(x) is continuous when a = 8
- f(x) is discontinuous for every value of a
- None of these
Correct answer: f(x) is continuous when a = 8
Solution
Left limit: lim_(x->0-) (1-cos4x)/x². Using (1-cosu)/u² -> 1/2 as u->0, with u=4x: = (16x²/2)/x²... more precisely = 2*sin²(2x)/x² = 2*4*(sin(2x)/2x)² -> 8. Right limit: sqrt(x)/(sqrt(16+sqrt(x))-4). Rationalise by multiplying by (sqrt(16+sqrt(x))+4): = sqrt(x)*(sqrt(16+sqrt(x))+4)/sqrt(x) = sqrt(16+sqrt(x))+4 -> sqrt(16)+4 = 8 as x->0+. Both limits equal 8, so a = 8 for continuity.
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