यदि `f(x)=|x-1| +|x+2|` जहा कोई भी वास्तविक संख्या है , तो सिद्ध कीजिए कि फलन `f(x),x=1` तथा `x =-2` पर संतत है ।

Explanation

`x=1` पर दक्षिण सीमा ( R.H.L)
`f ( x) = |x-1| + |x +2 |`
`f(x + h ) = |x + h - 1 | + | x + h + 2 | `
` f( 1 + h) = | 1 + h - 1 | + | 1+ h + 2 | `
` lim _( h to 0 ) f ( 1 = h ) = lim _( h to 0 ) [|h| + | 3 + h|] `
` = | 0 | + | 3 + 0 | = 3 `
` x = 1 ` पर वाम पक्ष सीमा ( L.H.L )
` f ( 1 + h ) = | 1 - h - 1 | + | - h + 2 | `
` f( 1 - h ) = | -h | + | 3 -h|`
`lim _ ( h to 0 ) f ( 1 - h ) =lim _ (h to 0)[ |h| + | 3- h |]`
` = |0| + | 3 - 0| = 3`
L.H.S = R.H.L
अतः `x=1` पर संतत है ।
` x = -2 ` पर दक्षिण सीमा ( R.H.L )
`f(x) = | x - 1 | + | x + 2| `
` f( -2 + h ) =| -2 + h - 1 | + | -2 + h + 2 | `
` lim _ ( h to 0 ) f ( -2 + h )= lim _ ( h to 0 ) [ | h - 3 | + | h | ] `
` = | -3 | + | 0 | = 3 `
` x = -2 ` पर वाम सीमा ( L.H .S)
` f(x) = | x - 1 | + | x + 2 | `
` f(-2 -h) = | -2 -h -1 | + | - 2 - h + 2| `
` = | -3 - h | + | -h|`
` = | 3 + h | + | h | `
` lim _( h to 0) f( -2-h) = lim _ ( h to 0 ) [ | 3 + h | + | h | ] `
` = 3`
अतः `x = -2 ` पर संतत है ।