We are asked to evaluate the following limit:

$\lim_{x \to 0} \frac{\sqrt[3]{8 + x} - 2}{x}$

Step 1: Substitute the hint $z = \sqrt[3]{8 + x}$

The hint suggests that we set $z = \sqrt[3]{8 + x}$, which implies:

$z^3 = 8 + x$

Rearranging this, we get:

$x = z^3 - 8$

Step 2: Rewrite the limit

Now, substitute $z$ and express the limit in terms of $z$. Notice that when $x \to 0$, we have $\sqrt[3]{8 + x} \to \sqrt[3]{8} = 2$, so $z \to 2$. Rewriting the limit:

$\lim_{x \to 0} \frac{\sqrt[3]{8 + x} - 2}{x} = \lim_{z \to 2} \frac{z - 2}{z^3 - 8}$

Step 3: Simplify the denominator

We can factor the denominator $z^3 - 8$ as a difference of cubes:

$z^3 - 8 = (z - 2)(z^2 + 2z + 4)$

Thus, the limit becomes:

$\lim_{z \to 2} \frac{z - 2}{(z - 2)(z^2 + 2z + 4)}$

Step 4: Cancel out $z - 2$

Since $z - 2$ appears in both the numerator and denominator, we can cancel it out (for $z \neq 2$):

$\lim_{z \to 2} \frac{1}{z^2 + 2z + 4}$

Step 5: Evaluate the limit

Now substitute $z = 2$ into the remaining expression:

$z^2 + 2z + 4 = 2^2 + 2(2) + 4 = 4 + 4 + 4 = 12$

Thus, the limit is:

$\frac{1}{12}$

Final Answer:

$\lim_{x \to 0} \frac{\sqrt[3]{8 + x} - 2}{x} = \frac{1}{12}$

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Would you like any further details on this process?

Related Questions:

1. How would you evaluate a similar limit involving square roots, like $\lim_{x \to 0} \frac{\sqrt{4 + x} - 2}{x}$?
2. How can you apply the concept of difference of cubes to other expressions?
3. What is the geometric meaning behind the cube root function and its behavior near a point?
4. How does L'Hopital's Rule apply to limits like this one?
5. Can you explain the difference between using a direct substitution and using factorization in limits?

Tip: When facing limits that result in the indeterminate form $\frac{0}{0}$, it's often helpful to factor the numerator and denominator, apply L'Hopital's Rule, or use a Taylor series expansion.