Question

Find the points of discontinuity of the function f, where `f(x) = {{:(x + 2",",  "if",  x ≥ 2),(x^2",",  "if",  x < 2):}`

Answer 1

`lim_(x -> 2^-) f(x) = lim_(x -> 2^-) x^2`

`lim_(x -> 2^-) f(x)` = 2² = 4  .......(1)

`lim_(x -> 2^+) f(x) = lim_(x -> 2^+) (x + 2)`

`lim_(x -> 2^+) f(x)` = 2 + 2 = 4  .......(2)

From equation (1) and (2), we get

`lim_(x -> 2^+) f(x) = lim_(x -> 2^+) f(x)`

∴ `lim_(x -> 2) f(x)` exist.

Let x₀ be an arbitrary point such that x₀ < 2.

Then `lim_(x -> x_0) f(x) =  lim_(x -> x_0) x^2`

`lim_(x -> x_0) f(x) =  x_0^2`

`f(x_0) = x_0^2`

∴ `lim_(x -> x_0) f(x) = f(x_0)`

For the point x₀ < 2, we have the limit of the function that exists and is equal to the value of the function at that point.

Since x₀ is an arbitrary point the above result is true for all x < 2.

∴ f(x) is continuous in `(- oo, 2)`.

Let x₀ be an arbitrary point such that x₀ > 2

Then `lim_(x -> x_0) f(x) =  lim_(x -> x_0) (x + 2)`

= x₀ + 2

`f(x_0) = x_0 + 2`

∴ `lim_(x -> x_0) f(x) = f(x_0)`

∴ For the point x₀ > 2, the limit of the function exists and is equal to the value of the function.

Since x₀ is an arbitrary point the above result is true for all x > 2.

∴ The function is continuous at all points of `(2, oo)`.

Hence the given function is continuous at all points of R.