Question

limits as x approaches 0 (2+x)^3 -8/x

Answer 1

What happens when you combine the fractions.

Answer 2

it doesnt work

Answer 3

I think this is meant to be written as:
\( \displaystyle \lim_{x \to 0} \left( \frac{(2+x)^3 - 8}{x} \right) \)

The trick here is to recognize the difference of cubes in the numerator. We can factor things of that form like so:
\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
The first cube is (2+x)^3. The second cube you do not directly see, but recognize that 8 = 2^3 by prime factorization!
\( \displaystyle \lim_{x \to 0} \left( \frac{(2+x)^3 - 2^3}{x} \right) \)
By factoring the difference of cubes, it becomes:
\( \displaystyle \lim_{x \to 0} \left( \frac{\color{red}{(\mathbf{2+x - 2})}((2+x)^2 + (2+x) \cdot 2 + 2^2)}{\mathbf{\color{red}{x}}} \right) \)

Notice that the two highlighted factors will become the same, and cancel each other. The remaining factor can be solved by substitution!