You may use the following formula such that:

`lim_(x-gt0) (a^x - 1)/x = ln a`

You need to form these fractions at numerator and denominator such that:

`lim_(x-gt0) ((7^x - 1)/x)/((4^x - 1)/x) * (x/x)`

Notice that you need to multiply `((7^x - 1)/x)/((4^x - 1)/x)` by (x/x) to preserve the initial fraction `(7^x - 1)/(4^x - 1).` `lim_(x-gt0) ((7^x - 1)/x)/((4^x - 1)/x) * (x/x) =` `(lim_(x-gt0) ((7^x - 1)/x))/(lim_(x-gt0)((4^x - 1)/x)) *lim_(x-gt0) (x/x)`

Considering the formula above, yields:

`lim_(x-gt0) ((7^x - 1)/x) = ln 7`

`` `lim_(x-gt0) ((4^x - 1)/x) = ln 4`

`lim_(x-gt0) (x/x) = 1`

`` `lim_(x-gt0) ((7^x - 1)/x)/((4^x - 1)/x) * (x/x) = ln7/ln4`

**Hence, evaluating the `lim_(x-gt0) (7^x - 1)/(4^x - 1)` yields: `lim_(x-gt0) (7^x - 1)/(4^x - 1) = ln7/ln4` .**

`lim_(x->0) (7^x-1)/(4^x-1)`

Since both `7^x-1` and `4^x-1` go to zero as x goes to zero I can use L'Hopital's rule to get

`lim_(x->0)(7^x-1)/(4^x-1)=lim_(x->0) (ln(7) 7^x)/(ln(4) 4^x)`

`lim_(x->0) 7^x = 1` and `lim_(x->0) 4^x = 1`

So `lim_(x->0) (7^x-1)/(4^x-1) = lim_(x->0) (ln (7) 7^x)/(ln (4) 4^x) = ln7/ln4~~1.403`

Hope this helps....