Question

How do you find the limit as x goes to 0 for the function `(3^x- 8^x)/ (x)`?

Answer 1

I haven't figured out how to find it algebraically. But here's a non-algebraic solution:

`lim_(xrarr0) (3^x-8^x) = 1-1=0` and, obviously, `lim_(xrarr0) x = 0`

So the limit has indeterminate form `0/0`.

Apply l'Hopital's Rule (it is not algebraic, it involves derivatives)

`d/dx(3^x-8^x) = 3^x ln3 - 8^x ln8` and, of course `d/dx(x) = 1`

So

`lim_(xrarr0) (3^x-8^x)/x = lim_(xrarr0) (3^x ln3-8^x ln8)/1 = ln3 - ln8`

Answer 2

Remembering the remarkable limit:

`lim_(xrarr0)(a^x-1)/x=lna`

than:

`lim_(xrarr0)(3^x-8^x)/x=lim_(xrarr0)(3^x-1+1-8^x)/x=`

`=lim_(xrarr0)[(3^x-1)/x-(8^x-1)/x]=ln3-ln8=ln(3/8)`.