Discuss the conjinuity of the functions at the points shown against them. If a function is discontinuous, determine whether the discontinuity is removable. In this case, redefine the function, so that it becomes continuous :
`{:(F(x)(4^(x)-e^(x))/(6^(x)-1)",for"x ne0),(=log((2)/(3))",for" x =0):}}at x =0.`

Explanation

Here, `f(0)=log((2)/(3))`
Now, `underset(x to 0)lim f(x) =underset(x to 0)lim(4^(x)-e^(x))/(6^(x)-1)`
`=underset(x to 0)lim((4^(x)-1)-(e^(x)-1))/(6^(x)-1)`
`=underset(x to 0)lim[(((4^(x)-1)-(e^(x)-1))/(x))/((6^(x)-x)/(x))]`
`=(underset(xto0)lim ((4^(x)-1)/(x))-underset(xto0)lim((e^(x)-1)/(x)))/(underset(xto 0)lim(6^(x)-1)/(x))`
`=(log4-loge)/(log6)=(log((4)/(e)))/(log6)`
`underset(xto0)limf(x)nef(0)`
`therefore f(x)` is discontinuous at `x=0.`
Here, `underset(x to 0)limf(x)` exists, but not equal to f(0). Hence, the discontinuity at `x=0` is removable and it can be removed by redefining the function as follows :
`f(x)=(4^(x)-e^(x))/(6^(x)-1),"for" x ne0`
`=(log ((4)/(e)))/(log6)'"for" x=0`