To solve this limit problem, let's examine the expression carefully:

$\lim_{{h \to 0}} \frac{2(1 + h)^4 - (1 + h)^3 + 2(1 + h)^2 - 3}{h}$

We will expand each term in the numerator and simplify before taking the limit.

Step 1: Expanding Each Term

1. Expand $(1 + h)^4$: $(1 + h)^4 = 1 + 4h + 6h^2 + 4h^3 + h^4$ So, $2(1 + h)^4 = 2 + 8h + 12h^2 + 8h^3 + 2h^4$.

2. Expand $(1 + h)^3$: $(1 + h)^3 = 1 + 3h + 3h^2 + h^3$ So, $-(1 + h)^3 = -1 - 3h - 3h^2 - h^3$.

3. Expand $(1 + h)^2$: $(1 + h)^2 = 1 + 2h + h^2$ So, $2(1 + h)^2 = 2 + 4h + 2h^2$.

Step 2: Substitute and Simplify

Now, substitute these expanded forms into the expression:

$2(1 + h)^4 - (1 + h)^3 + 2(1 + h)^2 - 3 = (2 + 8h + 12h^2 + 8h^3 + 2h^4) - (1 + 3h + 3h^2 + h^3) + (2 + 4h + 2h^2) - 3$

Combine like terms:

• Constant terms: $2 - 1 + 2 - 3 = 0$
• Terms with $h$: $8h - 3h + 4h = 9h$
• Terms with $h^2$: $12h^2 - 3h^2 + 2h^2 = 11h^2$
• Terms with $h^3$: $8h^3 - h^3 = 7h^3$
• Terms with $h^4$: $2h^4$

So the numerator simplifies to: $9h + 11h^2 + 7h^3 + 2h^4$

Step 3: Factor Out $h$

$\frac{9h + 11h^2 + 7h^3 + 2h^4}{h} = 9 + 11h + 7h^2 + 2h^3$

Step 4: Take the Limit as $h \to 0$

As $h \to 0$, the terms containing $h$, $h^2$, and $h^3$ will approach zero, leaving: $\lim_{{h \to 0}} \left(9 + 11h + 7h^2 + 2h^3\right) = 9$

Final Answer:

The limit is: $\boxed{9}$

This matches option A.

Would you like further details or have any questions about any of the steps?

Here are 5 related questions:

1. How do you handle higher powers in limits when $h$ approaches zero?
2. What are other methods to simplify expressions involving limits besides expansion?
3. Why does factoring out $h$ help in solving limits of this form?
4. How does L'Hôpital's Rule apply to similar limit problems?
5. What is the significance of higher-order terms (like $h^2, h^3$) as $h \to 0$?

Tip: When dealing with polynomial expansions in limits, always organize and simplify terms carefully to see what can be canceled or made negligible as the variable approaches zero.