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The problem given is:

\[
\lim_{{h \to 0}} \frac{f(1+h) - f(1)}{h} \quad \text{where} \quad f(x) = 2x^3 - 2
\]

This is the definition of the derivative of \( f(x) \) evaluated at \( x = 1 \). We will compute this step by step.

### Step 1: Find \( f(1+h) \)
Given \( f(x) = 2x^3 - 2 \), substitute \( x = 1 + h \):

\[
f(1+h) = 2(1+h)^3 - 2
\]

Now expand \( (1+h)^3 \):

\[
(1+h)^3 = 1 + 3h + 3h^2 + h^3
\]

So,

\[
f(1+h) = 2(1 + 3h + 3h^2 + h^3) - 2
\]
\[
f(1+h) = 2 + 6h + 6h^2 + 2h^3 - 2
\]
\[
f(1+h) = 6h + 6h^2 + 2h^3
\]

### Step 2: Find \( f(1) \)
Using \( f(x) = 2x^3 - 2 \), substitute \( x = 1 \):

\[
f(1) = 2(1)^3 - 2 = 2 - 2 = 0
\]

### Step 3: Substitute into the difference quotient
Now substitute \( f(1+h) \) and \( f(1) \) into the expression for the limit:

\[
\frac{f(1+h) - f(1)}{h} = \frac{(6h + 6h^2 + 2h^3) - 0}{h} = \frac{6h + 6h^2 + 2h^3}{h}
\]

Simplify the expression:

\[
\frac{6h + 6h^2 + 2h^3}{h} = 6 + 6h + 2h^2
\]

### Step 4: Take the limit as \( h \to 0 \)
As \( h \to 0 \):

\[
6 + 6h + 2h^2 \to 6
\]

Therefore, the value of the limit is:

\[
\boxed{6}
\]

Let me know if you'd like further details on any of the steps.

### Related Questions:
1. What is the general formula for the derivative using limits?
2. How do you expand a binomial expression like \( (1+h)^3 \)?
3. How would you calculate the derivative for \( f(x) = 2x^3 - 2 \) directly using differentiation rules?
4. What are the conditions under which a limit fails to exist?
5. How can the derivative be interpreted geometrically?

### Tip:
When computing limits, it's often helpful to simplify the expression first, especially when dealing with small values like \( h \to 0 \), as this can make the calculation easier.

lim (h→0) [f(1+h) - f(1)] / h, where f(x) = 2x^3 - 2

Learn how to calculate the limit lim (h→0) [f(1+h) - f(1)] / h, where f(x) = 2x^3 - 2. This problem demonstrates key calculus concepts, including limits and the definition of the derivative, with step-by-step explanations. Suitable for high school students in Grades 11-12.

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