`T h e v a l u e o flim_(x->0)(2-(256-7x)^(1//8))/((5x+32)^(1//5)-2)` is

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

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

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

If f(x) = (2 - (256 - 7x)^(1//8))/((5x + 32)^(1//5) - 2), x ne 0 , then for f to be continuous everywhere, f(0) is equal to :

If f(x)=(2-(256-7x)^(1//8))/((5x+32)^(1//5)-2) ( x ne 0) , then for f to be continuous on [-1,1] , f(0) is equal to

The value of lim_(x->0)((1+2x)/(1+3x))^(1/x^2)e^(1/x) is e^(5/2) b. e^2 c. e^(-2) d. 1

The value of lim_(x rarr0)((1+2x)/(1+3x))^((1)/(x^(2)))*e^((1)/(x)) is e^((5)/(2))b.e^(2)ce^(-2)d.1

Evaluate: lim_(x-0)((e^x-1)log_5(1+5x))/(sin^2x)

If f(x)=((256+ax)^((1)/(8))-2)/((32+bx)^((1)/(5))-2) is continuous at x=0 then the value of (a)/(b) is