निम्न फलानो की सांतत्यता की जाँच कीजिये –
(i) ` f (x) = {{:( (1)/(1 – e^(1//x))”,” x ne 0), ( 0″ “”,” x = 0):} ` पर
(ii) ` f ( x) = {{:( e^(1//x)”,”x ne 0),(0″ “”,” x = 0):} ` पर
(iii) ` f ( x ) = {{:( (1)/(1 + e^(1//x))”,” x ne0), ( 0″ “”,” x =0):} ` पर
(i) ` f (x) = {{:( (1)/(1 – e^(1//x))”,” x ne 0), ( 0″ “”,” x = 0):} ` पर
(ii) ` f ( x) = {{:( e^(1//x)”,”x ne 0),(0″ “”,” x = 0):} ` पर
(iii) ` f ( x ) = {{:( (1)/(1 + e^(1//x))”,” x ne0), ( 0″ “”,” x =0):} ` पर
( i) ` f( 0 – 0) = lim _ ( hto 0) f ( 0 – h) = lim _ ( hto 0) (1) /( 1 – e ^(-1//h))`
` = lim _ ( hto 0) ( 1)/(1 – (1 – ( 1) /(h) + ( 1 ) /(h ^(2)) * ( 1) /(2!) – …))`
` = lim _ ( hto 0) ( 1 ) /(((1)/(h) – (1)/(h^(2)*2 !) + …)) = (1) /((1)/(0) – ( 1)/(0*2!)) = ( 1)/(oo) = 0 `
इसी प्रकार ` f ( 0 + 0 ) = 0 `
और ` f ( 0 ) = 0 `
` rArr f ( 0 + 0) = f ( 0 – 0 ) = f (0) `