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Question
The following function has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it become continuous :
f(x) `{:(= (x^3 - 8)/(x^2 - 4)",", "for" x > 2),(= 3",", "for" x = 2),(= ("e"^(3(x - 2)^2 - 1))/(2(x - 2)^2) ",", "for" x < 2):}`
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Solution
f(2) = 3 ...(Given)
`lim_(x -> 2^-) "f"(x) = lim_(x -> 2^-) (x^3 - 8)/(x^2 - 4)`
= `lim_(x -> 2^-) (x^3 - 2^3)/(x^2 - 2^2)`
= `lim_(x -> 2^-) ((x - 2) (x^2 + 2x + 4))/((x - 2)(x + 2))`
= `lim_(x -> 2^-) (x^2 + 2x + 4)/(x + 2)`
= `(lim_(x -> 2^-) (x^2 + 2x + 4))/(lim_(x -> 2^-) (x + 2))`
= `((2)^2 + 2(2) + 4)/(2 + 2)`
= `12/4`
= 3
`lim_(x -> 2^+) "f"(x) = lim_(x -> 2^+) ("e"^(3(x - 2)^2) - 1)/(2(x - 2)^2)`
Put x – 2 = h
∴ x = 2 + h
As x → 2, h → 0
∴ `lim_(x -> 2^+) "f"(x) = lim_("h" -> 0) ("e"^(3"h"^2) - 1)/(2"h"^2)`
= `1/2 lim_("h" -> 0) ("e"^(3"h"^2) - 1)/(3"h"^2) xx 3`
= `1/2 xx 1 xx 3 ...[(because "h" -> 0 therefore "h"^2 -> 0),(and lim_(x -> 0) ("e"^x - 1)/x = 1)]`
= `3/2`
∴ `lim_(x -> 2^-) "f"(x) ≠ lim_(x -> 2^+) "f"(x)`
∴ `lim_(x -> 2) "f"(x)` does not exist
∴ f(x) is discontinuous at x = 2
This discontinuity is irremovable.
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