If the function `f(x)=((4^(sin x)-1)^(2))/(x*log(1+2x))`, for `x!=0` is continuous at `x=0`, find `f(0)`.

Explanation

Given,`f(x)=((4^(sin x)-1)^(2))/(x*log(1+2x))`
`underset(x rarr 0)limf(x)=underset(x rarr 0)lim ((4^(sinx)-1)^(2))/(x*log(1+2x))`
`=underset(xrarr0)lim .(((4^(sin x)-1)/(sin x))^(2)xx(sin^(2)x)/(x^(2)))/((log(1+2x))/(2x)xx2) [:'x rarr 0 x != 0, sin x != 0]`
`=((underset(x rarr 0)lim(4 ^(sin x)-1)/(sin x))^(2)xx(underset(xrarr0)lim(sinx)/(x))^(2))/(2underset(x rarr 0)lim (log(1+2x))/(2x))`
`=((log4)^(2)xx 1^(2))/(2xx1)[(.:.underset(x rarr 0)lim (a^(x)-1)/(x)=log a.),(underset(x rarr 0)lim (1+t)/(t)=1.)]`
`:.underset(xrarr0)lim f(x)=((log4)^(2))/(2)`
Now, `f(x)` is continuous at `x=0`,
`rArr f(0)=underset(xrarr0)lim f(x)`
`:.f(0)=((log4)^(2))/(2)`