फलन f(x) की x = 0,1 पर सांतत्य का परिक्षण कीजिए जब
` f(x)={:(1-2x, " यदि " x lt 0),(1, " यदि "0 le x lt 1),(2x-1, " यदि " x ge 1):}`

Explanation

(i) x =0 पर सांतत्य
`f (0) = 1`
` R.H.L = lim _(x to 0+ 0) f ( x) = lim _(h to 0) f ( 0 + h ) `
` = lim _ (h to 0) (1) = 1 `
`L.H.S = lim _(x to 0-0) f(x) = lim _(h to 0 ) f (0-h) `
` = lim _(h to 0) ( 1 + 2h) = 1 `
` :. L.H.S = f(0) = R.H.L .`
अतः फलन` x=0` पर संतत है ।
(ii) ` x= 1 पर सांतत्य `
` f(1) = (2 xx 1 - 1) = 1,`
`L.H.S = lim_(x to 1 - 0) f(x) = lim_(h to 0) f(1 - h) `
` = lim _( h to 0 ) (1) = 1 `
`R.H.L = lim _( x to 1 + 0) f(x) = lim _( h to 0 ) f ( 1 + h ) `
` = lim _( h to 0) { 2 ( 1 + h 0 - 1 } `
` = lim _( h to 0 ) { 1 + 2h} `
=1
`:. L.H.L = f(1) = R.H.L `
अतः फलन `x= 1 ` संतत है ।