#### 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

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

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

`=(log4-loge)/log6`

`=log(4/e)/log6`

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

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

`=(log4-loge)/log6`

`=log(4/e)/log6`

LHL = RHL at x = 0.

`But f(0) != lim_(x->0)f(x).`

Hence, the given function has removable discontinuity at x = 0.

To remove the discontinuity, we define `f(0)=log(4/e)/log6`

So the revised function becomes

`f(x)={((4^x-e^x)/(6^x-1), x!=0),(log(4/e)/log6,x=0) :}`