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प्रश्न
Discuss the continuity of f at x = 1
Where f(X) = `[ 3 - sqrt ( 2x + 7 ) / ( x - 1 )]` For x ≠ 1
= `-1/3` For x = 1
उत्तर
Given f(1) = `-1/3` (i)
Now `lim_( x -> 1 ) = lim_( x -> 1 ) ([ 3 - sqrt (2x + 7) ]/[ x - 1 ])`
= `lim_( x -> 1 ) [ 3 - sqrt( 2x + 7 )]/[ x - 1 ] xx [ 3 + sqrt( 2x + 7 ) ]/[ 3 + sqrt( 2x + 7 )]`
= `lim_( x -> 1) [ 9 - (2x + 7) ]/[( x - 1)( 3 + sqrt( 2x + 7 ))]`
= `lim_( x -> 1) [ 2 - 2x ]/[( x - 1)( 3 + sqrt( 2x + 7)]`
= `lim_( x -> 1) [ -2( x - 1 )]/[( x - 1)( 3 + sqrt( 2x + 7)]`
= `lim_( x -> 1) [ -2]/[3 + sqrt( 2x + 7]`
= `[ - 2 ]/[ 3 + sqrt 9] = -2/6 = -1/3`
`therefore lim_( x -> 1) f(x) = -1/3` (ii)
From (i) and (ii), `lim_( x -> 1) f(x) = f(1)`
∴ f is continuous at x = 1.
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