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The value of `f(0),` so that the function `f(x)=((27-2x)^2-3)/(9-3(243+5x)^(1//5)-2)(x!=0)` is continuous, is given `2/3` (b) 6 (c) 2 (d) 4

A

`(2)/(3)`

B

`6`

C

`2`

D

`4`

Text Solution

Verified by Experts

The correct Answer is:
C
Since f (x) is continuous at x = 0, therefore
`f(0) = underset(xrarr0)lim f (x) = underset(xrarr0)lim ((27-2x)^(1//3)-3)/(9-3(243+5x)^(1//5))`
`("from"(0)/(0))`
`=underset(xrarr0)lim((1)/(3)(27-2x)^(-2//3)(-2))/(-(3)/(5)(243+_5x )^(-4//5)(5))=2`
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