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Question
Show that the function `{{:((x^3 - 1)/(x - 1)",", "if" x ≠ 1),(3",", "if" x = 1):}` is continuous om `(- oo, oo)`
Solution
`{{:((x^3 - 1)/(x - 1)",", "if" x ≠ 1),(3",", "if" x = 1):}`
Clearly, the given function f(x) is defined at all points of R.
Case (i) Let x0 ∈ `(- oo, 1)` then
`lim_(x -> x_0) f(x) = lim_(x -> x_0) (x^3 - 1)/(x - 1)`
= `(x_0^3 - 1)/(x_0 - 1) x_0 ≠ 1`
`f(x_0) = (x_0^3 - 1)/(x_0 - 1) x_0 ≠ 1`
∴ `lim_(x -> x_0) f(x) = f(x_0)`
f(x) is continuous at x = x0.
Since x0 is arbitrary f(x) is continuous at all points of `(-oo, 1)`.
Case (ii) Let x0 ∈ `(1, oo)` then
`lim_(x -> x_0) f(x) = lim_(x -> x_0) (x^3 - 1)/(x - 1)`
= `(x_0^3 - 1)/(x_0 - 1) x_0 ≠ 1`
`f(x_0) = (x_0^3 - 1)/(x_0 - 1) x_0 ≠ 1`
∴ `lim_(x -> x_0) f(x) = f(x_0)`
f(x) is continuous at x = x0.
Since x0 is arbitrary f(x) is continuous at all points of `(1, oo)`.
Case (iii) Let x0 = 1 then
`lim_(x -> 1) f(x) = lim_(x -> 1) 2` = 2
`f(1)` = 2
∴ `lim_(x -> 1) f(x) = f"'"(1)`
Hence, f(x) is continuous at x = 1.
Using all the three cases, we have f(x) is continuous at all the points of R.
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