Find the limit as x approaches 2 of (1 + (x cubed - 8)/(4x - 8)) raised to the power 2x.

Step-by-step solution

1. 1

   Diagnose the form before choosing a method. At $x=2$ the fraction $\frac{x^3-8}{4x-8}$ is $\frac00$, so the base is not obviously $1$ - this may or may not be a $1^\infty$ situation. Simplify the inner fraction first and the form will declare itself.

2. 2

   Factor numerator and denominator. The numerator is a difference of cubes: $x^3-8 = x^3-2^3 = (x-2)(x^2+2x+4)$. The denominator is $4x-8 = 4(x-2)$. The common factor $x-2$ is nonzero on the approach (we never evaluate at $2$), so it cancels: $\frac{x^3-8}{4x-8} = \frac{x^2+2x+4}{4}.$

3. 3

   Rebuild the base. $1+\frac{x^2+2x+4}{4} = \frac{4+x^2+2x+4}{4} = \frac{x^2+2x+8}{4}.$

4. 4

   Evaluate the base and the exponent separately. Both are now continuous at $x=2$. Base: $\frac{4+4+8}{4} = \frac{16}{4} = 4$. Exponent: $2x = 4$. So the form is $4^4$, an ordinary determinate limit - not $1^\infty$, and no exponential/logarithm trick is required.

5. 5

   Compute the value. $\lim_{x\to2}\left(1+\frac{x^3-8}{4x-8}\right)^{2x} = 4^4 = 256.$

6. 6

   Confirm numerically. At $x=2.0001$ the expression evaluates to $256.109$, and at $x=1.9999$ it is just under $256$ - closing in on $256$ from both sides.