It the function `f(x) = ((5^(sinx) - 1)^(2))/(x log (1 +2 x)) `for `x ne 0` is continous at x = 0 find f(0).

Explanation

Since `f(x) = ((5^(sinx) - 1)^(2))/(x log (1+2x))` is continous at x = 0
`therefore" "f(0) = underset(x to 0)(lim)f(x)`
` = underset(x to0)(lim)((5^(sinx) -1)^(2))/(x log (1+2x))`
` = underset(x to 0)(lim)[((((5^(sinx)-1)^(2))/(sinx))((sin^(2)x)/(x^(2))))/((xlog (1+2x))/(x^(2)))]`
` = (underset(x to0)(lim)((5^(sinx) - 1)/(sinx))^(2) underset(x to0)(lim)((sinx)/(x))^(2))/(2underset(xto0)(lim)(1)/(2x)log(1+2x)^(1//2x))`
`= ((log 5)^(2))/(2log e)`
`f(0) = ((log5)^(2))/(2)" "[because log e = 1]`