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प्रश्न
Let f(x) = `{{:((1 - cos 4x)/x^2",", "if" x < 0),("a"",", "if" x = 0),(sqrt(x)/(sqrt(16) + sqrt(x) - 4)",", "if" x > 0):}`. For what value of a, f is continuous at x = 0?
उत्तर
Here f(0) = a Left hand limit of f at 0 is
`lim_(x -> 0^-) "f"(x) = lim_(x -> 0^-) (1 - cos 4x)/x^2`
= `lim_(x -> 0^-) (2sin^2 2x)/x^2`
= `lim_(2x -> 0^-) 8((sin 2x)/2x)^2`
= 8(1)2
= 8.
And right hand limit of f at 0 is
`lim_(x -> 0^+) "f"(x) = lim_(x -> 0^+) sqrt(x)/(sqrt(16 + sqrt(x)) - 4)`
= `lim_(x - 0^+) (sqrt(x)(sqrt(16 + sqrt(x)) + 4))/((sqrt(16 + sqrt(x)) + 4)(sqrt(16 + sqrt(x)) - 4))`
= `lim+_(x -> 0^+) (sqrt(x)(sqrt(16 + sqrt(x)) + 4))/(16 + sqrt(x) 16)`
= `lim_(x -. 0^+) (sqrt(16 + sqrt(x)) + 4)`
= 8
Thus, `lim_(x -> 0+) "f"(x) = lim_(x -> 0^+) "f(x)` = 8.
Hence f is continuous at x = 0 only if a = 8
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