फलन ` f ( x ) ` एक अंतराल ` [0, 1 ] ` में निम्नवत परिभाषित है -
` f ( 0 ) = 0`
` f ( x ) = ( 1) /(2) - x ` यदि ` 0 lt x lt (1) /(2) `
` f ( ( 1 ) /(2)) = ( 1 ) /(2) `
` f ( x) = (2) /(3) - x ` यदि ` ( 1 ) /(2) lt x lt 1 `
` f ( 1 ) = 1 `
वह बिंदु ज्ञात कीजिये जिन पर फलन असतत है |

Explanation

(i) ` x = 0 ` पर ` f ( 0 ) = 0 `
फलन की निम्न सीमा पर दक्षिण पक्ष सीमा
माना ` x = 0+ h ` यदि ` x = 0 rArr h = 0 `
` therefore lim_ ( x to 0 ) rArr lim_ ( hto 0 ) `
` therefore lim _ ( x to 0 ^(+)) f ( x ) = lim _ ( h to 0 ) f ( 0 + h ) `
` = lim _ ( h to 0) [ ( 1) /(2) - (0 + h) ] `
` ( because f (x) = (1 ) /(2) - x ` का प्रयोग किया जायेगा |)
` = (1)/(2) ne f (0 ) `
` therefore x = 0 ` पर फलन असतत है |
(ii) ` x = (1) /(2) ` पर ` f ( (1)/(2)) = (1)/(2) `
फलन की दक्षिण पक्ष सीमा
माना ` x =(1)/(2) + h ` यदि ` x = (1)/(2) rArr h = 0 `
` therefore lim _ ( x to 1//2) rArr lim _ ( h to 0 ) `
` therefore lim _ ( x to ((1)/(2))^(+)) f ( x ) = lim _ ( h to 0 ) f ((1 ) /(2)+ h) `
` = lim_ ( h to 0 ) [ ( 2) /(3) - ( ( 1 ) /(2) + h)] `
` ( because x gt ( 1 ) /(2) therefore f ( x ) = (2) /(3) - x ` का प्रयोग किया जायेगा | )
` = lim _ ( h to 0 ) [ (2)/(3) - (1) /(2) - h ] `
` = ( 2)/(3) - (1)/(2) - 0 = (1)/(6) ne f (( 1 ) /(2)) `
` therefore x = (1)/(2) ` पर फलन असतत है |
(iii) फलन की उच्च सीमा ` x = 1 ` पर ` f ( 1 ) = 1 `
फलन की वाम पक्ष सीमा
माना ` x = 1 - h ` यदि ` x = 1 ` तब h = 0
` therefore lim _ ( x to1 ) rArr lim _ ( h to 0 ) `
` therefore lim _ ( x to 1 ^-) f (x ) = lim _ ( h to 0 ) f ( 1 - h ) `
` = lim _ ( h to 0 ) [ (2)/(3) - ( 1 - h)] `
` ( because x lt 1 therefore f ( x ) = (2)/(3) - x ` का प्रयोग किया जायेगा |)
` = lim _ ( h to 0 ) [ ( 2 ) /(3) - 1 + h ] `
` = lim _ ( h to 0 ) [ - (1) /(3) + h ] = - (1)/(3) + 0 = - (1)/(3) `
अतः फलन x =1 पर असतत है |