Question

Show that the function `{{:((x^3 - 1)/(x - 1)",",  "if"  x ≠ 1),(3",",  "if"  x = 1):}` is continuous om `(- oo, oo)`

Answer 1

`{{:((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.