Question

How do you find the limit `lim (3-sqrt3^x)/(9-3^x)` as `x->2`?

Answer 1

`1/6`

Explanation:

You can factorize the denominator by the rule:

`a^2-b^2=(a-b)(a+b)`

Then

`9-3^x=(3-sqrt3^x)(3+sqrt3^x)`

Let's substitute and simplify:

`lim_(x->2)(3-sqrt(3^x))/(9-3^x)=lim_(x->2)cancel(3-sqrt(3^x))/(cancel((3-sqrt3^x))(3+sqrt3^x))=1/(3+cancelsqrt3^cancel2)=1/6`