यदि
` f ( x ) = {{:( (sin (a + 1) x + 2 sin x ) /( x ),",", xlt 0 ), ( 2, ",", x = 0 ), ( (sqrt( 1 +bx)- 1 )/( x ),",", x gt 0 ):}`
` x = 0 ` पर सतत है, तब a व b के मान ज्ञात कीजिये |

Explanation

दिया है,

` f ( x ) = {{:( (sin (a + 1) x + 2 sin x ) /( x ),",", xlt 0 ), ( 2, ",", x = 0 ), ( (sqrt( 1 +bx)- 1 )/( x ),",", x gt 0 ):}`

दिया है कि फलन ` f ( x ) , x = 0 ` पर सतत है |

` therefore ` (बायाँ पक्ष ) `""_(x = 0 ) ` = ( दायाँ पक्ष) `""_(x =0) ` = ` f ( 0 ) ` ...(1)

अब, बायाँ पक्ष = ` lim_ ( x to 0^-) f (x) `

` = lim _ ( hto 0) f ( 0 - h ) `

` [ x = 0 - h ` रखने पर, जब ` x to 0 ` , तब ` h to 0 ] `




` = lim _ ( hto 0) ( - sin ( a + 1 ) h - 2 sin h )/( - h) `

`" " [ because sin ( -theta) = -sin theta ] `

` = lim _ ( h to 0) ( sin ( a + 1 ) h+ 2 sin h ) /( h) `

` = lim _ ( hto 0 ) ( sin ( a + 1 ) h ) /( h ) + lim _ ( hto 0) ( 2 sin h ) /( h ) `

` = lim _ ( hto 0 ) ( sin ( a+ 1 ) h ) /( ( a + 1 ) h) xx ( a + 1 ) + 2 lim_ ( hto0 ) ( sin h ) /( h) `

` = 1xx ( a + 1 ) + 2 xx 1`

` [ because lim _ ( x to 0) ( sin x ) /(x) = 1 ] `

` = a + 1 + 2 = a + 3 `

तथा दायाँ पक्ष ` = lim _ ( xto 0^+) f (x) `


` = lim _ ( hto 0) f ( 0 + h) `

` [ x = 0 + h` रखने पर, यदि ` x to 0 ` तब ` h to 0 ] `

` = lim _ ( hto 0) ( sqrt ( 1 + b ( 0 + h) ) - 1 ) /( 0 + h )`

` = lim_ ( hto 0 ) ( sqrt ( 1 + bh) - 1) /( h ) `



[अंश व हर में ` sqrt ( 1 + bh ) + 1 ` से गुणा करने पर ]

` = lim _ ( hto 0 ) (( 1 + bh ) - 1 ) /( h ( sqrt ( 1 + bh ) + 1 ))`

` = lim _ ( hto 0) ( bh ) /(hsqrt( 1 +bh ) + 1 ) `

` = lim _ (h to 0) ( b) /(( sqrt ( 1 + bh ) + 1 )) = (b) /(sqrt ( 1+ 0 ) + 1 ) = (b) /(2) `

समीकरण (1 ) से,

` a + 3 = ( b) /(2) = 2 `

` rArr a + 3 = 2 ` तथा `(b)/(2) = 2 `

` rArr a = - 1 ` तथा ` b = 4 `