#### Question

Discuss the continuity of the following functions. If the function have a removable discontinuity, redefine the function so as to remove the discontinuity

`f(x)=(4^x-e^x)/(6^x-1)` for x ≠ 0

`=log(2/3) ` for x=0

#### Solution

f(0) = log `(2/3)` .....Given .....(1)

`lim_(x->0^-f(x))=lim_(x->0^-)(4^x-e^x)/(6^x-1)`

`= lim_(x->0) ((4^x - 1) - (e^x - 1))/(6^x - 1) `

`lim_(x->0) ((4^x - 1)/x - (e^x - 1)/x)/((6^x - 1)/x)` .....[x → 0 , x ≠ 0]

`lim_(x->0) (lim_(x->0)(4^x - 1)/x - lim_(x->0)(e^x - 1)/x)/(lim_(x->0)(6^x - 1)/x)`

`=((log4) - 1)/log6` .......`[because lim_(x->0) (a^x - 1)/x = log e]`

`therefore lim_(x->0) f(x) = ((log4) - loge)/log6`

`therefore lim_( x ->0) = [log4]/[loge.log6]`

`therefore lim_( x ->0) = [log4]/[1.log6]`

`therefore lim_( x ->0) = log(2/3)`

From (1) and (2) , lim_(x->0) f(x) ≠ f(0)

∴ f is discontinuous at x = 0

Here lim_(x->0) 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 ≠ 0

`=log(2/3) ` for x=0