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Discuss the continuity 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 .`

Text Solution

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`f(0)=log((2)/(3))" "...("Given")...(1)`
`underset(x to0)limf(x)=underset(x to0)lim(4^(x)-e^(x))/(6^(x)-1)`
`=underset(x to0)lim((4^(x)-1)-(e^(x)-1))/(6^(x)-1)`
`=underset(xto0)lim(((4^(x)-1)/(x))-((e^(x)-1)/(x)))/(((6^(x)-1)/(x)))" "...[x to0, x ne0]`
`=(underset(xto0)lim(4^(x)-1)/(x)-underset(xto0)lim(e^(x)-1)/(x))/(underset(xto0)lim(6^(x)-1)/(x))`
`=((log4)-1)/(log6)" "...[becauseunderset(xto0)lim(a^(x)-1)/(x)=loga]`
`thereforeunderset(xto0)limf(x)=((log4)-1)/(log6)" "...(2)`
From (1) and (2), `underset(xto0)lim f(x) ne f(0)`
`thereforef` is discontinous at `x=0.`
Here `underset(xto0)lim` f(x) exists but not equal to f(0). Hence, the discontinuity at `x=0` is removable and can be removed by redefining the function as follows :
`f(x)=(4^(x)-e^(x))/(6^(x)-1), "for" x ne0`
`=((log4)-1)/(log6),"for"x=0.`
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