Is the function f defined by
`f(x)={{:(x, if x le 1),(5, if x gt 1):}`
continuous at x = 0? At x = 2 ?

Explanation

`f(x)={{:(x, if x le 1),(5, if x gt 1):}`
at x = 0
Let x = 0 - h
`rArr 0 - h to 0`
`rArr h to 0`
L.H.L = ` underset (X to 0^(-))(lim)f(x)`
` underset (h to 0)(lim)f(0 - h)`
` underset (h to 0)(lim)(0 -h) = 0 `
f(0) = 0
Let x = 0 + h
`rArr 0 - h to 0 `
`rArr h to 0 `
R.H.L. = ` underset (x to 0^(+))(lim)f(x)`
` underset (h to 0)(lim)f(0 + h)`
` underset (h to 0)(lim)(0 + h) = 0 `
`:'` L.H.L. = f(0) = R. H.L.
`:.` f(x) is continuous at x = o.
at x = 1
Let x = 1 - h
`rArr 1 - h to 1`
`rArr h to 0 `
` f(1) = 1 `
L.H.L. = ` underset (x to 0^(-))(lim)f(x) `
` underset (h to 0)(lim)f(1 - h) `
` underset (h to 0)(lim)f(1 - h)= 1-0=1`
R.H.L. = ` underset (x to 1^(+))(lim)f(x)`
Let x = 1 + h
`rArr 1+ hto 1`
`rArr h to 0`
` underset (h to 0)(lim)f(1 + h)`
` underset (h to 0)(lim)5 =5`
`:'` L.H.L. `ne` R.H.L.
`:.` f(x), is not continuous at x = 1 ,brgt at x = 2
Let x = 2 - h
`rArr 2 - to 2`
`rArr h to 0`
f(2) = 5
L.H.L = ` underset (X to 2^(2))(lim)f(x)`
` underset (h to 0)(lim)f(2-h)`
` underset (h to 0)(lim)5=5`
R.H.L. = ` underset (x to 2^(+))(lim)f(x)`
Let x = 2 + h
`rArr 2 + h to 2`
`rArr h to 0 `
` underset (h to 0)(lim)f(1 + h)`
` underset (h to 0)(lim)5=5`
`:'` L.H.L. = f(2) = R.H.L.
`:.` f(x) is continuous at x = 2