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प्रश्न
Show that following function have continuous extension to the point where f(x) is not defined. Also find the extension :
f(x) = `(x^2 - 1)/(x^3 + 1)` for x ≠ – 1
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उत्तर
f(x) = `(x^2 - 1)/(x^3 + 1)`; x ≠ – 1
Here f(– 1) has not been defined.
Consider
`lim_(x -> -1) "f"(x) = lim_(x -> - 1) ((x^2 - 1)/(x^3 + 1))`
= `lim_(x -> - 1) ((x + 1)(x - 1))/((x + 1)(x^2 - x + 1))`
= `lim_(x -> - 1) (x - 1)/(x^2 - x + 1)` ...[∵ x → – 1, ∴ x ≠ – 1, ∴ x + 1 ≠ 0]
= `(-1 - 1)/((-1)^2 - (-1) + 1)`
= `-2/3`
Thus `lim_(x -> -1)` f(x) exists but f(– 1) is not defined.
∴ f(x) has a removable discontinuity at x = – 1
∴ The extension of the original function is
f(x) = `{:(= (x^2 - 1)/(x^3 + 1), ; x ≠ – 1),(= -2/3, ; x = - 1):}`
f(x) is continuous at x = `-2/3`
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