We are tasked with evaluating:

$\lim_{x \to 0} \frac{\cos(x) - 1 + \frac{1}{2}x^2}{2x^4}.$

Step 1: Analyze the numerator

The numerator is $\cos(x) - 1 + \frac{1}{2}x^2$. Using the Taylor expansion of $\cos(x)$ about $x = 0$: $\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^6),$ we substitute this into the numerator: $\cos(x) - 1 + \frac{1}{2}x^2 = \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) - 1 + \frac{1}{2}x^2.$

Simplify the terms: $\cos(x) - 1 + \frac{1}{2}x^2 = -\frac{x^2}{2} + \frac{1}{2}x^2 + \frac{x^4}{24} = \frac{x^4}{24}.$

Thus, the numerator simplifies to $\frac{x^4}{24}$.

Step 2: Write the limit in simplified form

The denominator is $2x^4$. Substituting the simplified numerator, the limit becomes: $\lim_{x \to 0} \frac{\frac{x^4}{24}}{2x^4}.$

Simplify the fraction: $\frac{\frac{x^4}{24}}{2x^4} = \frac{1}{24 \cdot 2} = \frac{1}{48}.$

Final Answer:

$\lim_{x \to 0} \frac{\cos(x) - 1 + \frac{1}{2}x^2}{2x^4} = \frac{1}{48}.$

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Would you like a detailed explanation of the Taylor expansion or any other step?
Here are 5 related questions to expand on this:

1. How do you compute higher-order terms in a Taylor series for trigonometric functions?
2. What is l’Hospital’s Rule, and how can it be applied to indeterminate forms?
3. Can Taylor expansions always simplify limits, and if not, what are alternatives?
4. What are common patterns of limits involving trigonometric functions and their series?
5. How does the order of approximation in Taylor expansions affect accuracy?

Tip: Always check for simplifications like Taylor expansions before applying l’Hospital’s Rule—it often avoids unnecessary derivatives!