The value of `f(0)` so that the function `f(x)=(2-(256-7x)^(1/8))/((5x+32)^(1//5)-2),x!=0` is continuous everywhere, is given by `-1` (b) 1 (c) 26 (d) none of these

Explanation

`f(0)=(2-2)/(2-2)=0/0`

`l=lim_(x->o) f(x)=lim_(x->o) 2-(256-7x)^(1/8)/((5x+32)(1/5)-2)`

`l`hospital rule

`l=lim_(x->o) (0-1/8(256-7x)^(1/8-1)(-2))/((1/5)(5x+32)^(1/5-1)-0)`

`l=lim_(x->o)((7/8)(256-7x)^(-7/8))/(5x+3x)^(-4/5)`

`l=lim_(x->o)((7/8(256)^-(7/8))/(32)^-(4/5))`

`l=lim_(x->o)((7/8(256)^-(1/8))^-7)/((32)^-(1/5))^-4`

`=7/8(2)^-7/(2)^-4`

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