a और c के किस मान के लिए, बिंदु ` x = 0 ` पर दिया गया फलन ` f ( x ) ` सतत है ?
` f ( x) {{:( ( sin ( a + 1 ) x + sinx ) /( x ) ",",, "यदि" x lt 0 ) , ( c",",, "यदि" x = 0 ) , ( (sqrt ( x + b x ^(2 ) ) - sqrt x ) /( b x ^(3//2))",",, "यदि" x gt 0 ) :} `

Explanation

` f ( x ) ` की परिभाषा के अनुसार
` f ( 0 - 0 ) = lim _ ( h to 0 ) f ( 0 - h ) `
` = lim _ ( hto 0 ) ( sin ( a + 1 ) ( - h ) + sin ( - h ) ) /( - h ) `
` = lim _ ( hto 0) (-{ sin ( a + 1 ) h + sin h })/( - h ) `
` = lim_ ( hto 0 ) ( sin ( a + 1 ) h + sin h ) /( h ) `
` = lim _ ( hto 0 ) ( 2sin ((a ) /(2)+ 1 ) h * cos ""(ah)/(2))/(h) `

` = 2 * lim _ ( h to 0 ) { (sin ( ( a ) /(2) + 1 ) h ) /( ((a ) /(2) +1) h ) (( a ) /(2)+1) * cos ""(ah)/(2)}`
` = 2 ((a ) /(2) + 1) lim _ ( h to 0 ) ( sin ((a) /(2) + 1 ) h ) /( ((a) /(2) + 1) h) * lim_ ( hto 0 ) cos ""(ah)/(2) `
` = ( a + 2) xx 1 xx 1 = ( a + 2 ) `
और ` f ( 0 + 0 ) = lim _ ( hto 0 ) f ( 0 + h ) `
` = lim _ ( hto 0 ) { ([ sqrt ( h + bh^2) - sqrt h ]) /(bh^(3//2)) xx ([sqrt ( h + bh ^(2)) + sqrt h])/( [ sqrt ( h +bh^(2)) + sqrth])} `
` = lim _ ( hto 0) (1)/((sqrt(1+bh ) + 1 )) = (1)/(2) `
अब बिंदु ` x = 0 ` पर ` f ( x ) ` की सततता के लिए
` f ( 0 ) = f ( 0 +0) = f ( 0 - 0 ) `
` rArr c = a + 2 = (1)/(2) `
` rArr a = ( - 3 )/(2) ` और ` c = ( 1 ) /(2) `