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प्रश्न
Discuss continuity of f(x) =`(x^3-64)/(sqrt(x^2+9)-5)` For x ≠ 4
= 10 for x = 4 at x = 4
उत्तर
`f(4) = 10`
`lim_(x→4)f(x)=lim_(x→4)(x^3-64)/(sqrt(x^2+9)-5)`
`=lim_(x→4)(x^3-64)/(sqrt(x^2+9)-5)`
`lim_(x→4)(x^3-4^3)/(sqrt(x^2+9)-5)xx(sqrt(x^2+9)+5)/(sqrt(x^2+9)+5)`
`Lim_(x→4)((x^3-4^3)(sqrtx^2+9+5))/((sqrtx^2+9)^2-(-5)^2)`
`Lim_(x→4) ((x-4)(x^2+4x+16)(sqrt(x^2+9)+5))/(x^2+9-25)` .......`[a^3 – b^3 = (a – b) (a^2 + ab + b^2))`
`Lim_(x→4) ((x-4)(x^2+4x+16)(sqrt(x^2+9)+5))/(x^2-16)`
`Lim_(x→4) ((x-4)(x^2+4x+16)(sqrt(x^2+9)+5))/((x-4)(x+4)`)
`Lim_(x→4) ((x^2+4x+16)(sqrt(x^2+9)+5))/(x+4)`
`(((4)^2+4(4)+16)(sqrt(4^2+9)+5))/(4+4)`
`((16+16+16)(sqrt(16+9)+5))/8`
`((16+16+16)(sqrt(25)+5))/8`
`((16+16+16)(5+5))/8`
`(48x10)/8`
`f(x)lim_(x→4)=60`
`lim_(x→4) f(x)≠f(4)`
∴ f (x) is not continuous at x = 4.
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