Let's solve the two limits step by step.

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

Step 1: Taylor expansion of $\cos x$: $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^6)$ Substitute this into the numerator: $\cos x - 1 + \frac{1}{2}x^2 = \left(1 - \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^6)\right) - 1 + \frac{1}{2}x^2$ Simplify: $\cos x - 1 + \frac{1}{2}x^2 = -\frac{x^2}{2} + \frac{1}{2}x^2 + \frac{x^4}{24} + \mathcal{O}(x^6) = \frac{x^4}{24} + \mathcal{O}(x^6)$

Step 2: Divide the numerator by $x^4$: $\frac{\cos x - 1 + \frac{1}{2}x^2}{x^4} = \frac{\frac{x^4}{24} + \mathcal{O}(x^6)}{x^4} = \frac{1}{24} + \mathcal{O}(x^2)$

Step 3: Take the limit as $x \to 0$: $\lim_{x \to 0} \frac{\cos x - 1 + \frac{1}{2}x^2}{x^4} = \frac{1}{24}.$

(ii) $\lim_{n \to \infty} \frac{n^6 + n^4 - 1000}{7n^6 + n^5 - 1}$

Step 1: Factor $n^6$ from numerator and denominator:

In the numerator: $n^6 + n^4 - 1000 = n^6\left(1 + \frac{1}{n^2} - \frac{1000}{n^6}\right).$

In the denominator: $7n^6 + n^5 - 1 = n^6\left(7 + \frac{1}{n} - \frac{1}{n^6}\right).$

The fraction becomes: $\frac{n^6\left(1 + \frac{1}{n^2} - \frac{1000}{n^6}\right)}{n^6\left(7 + \frac{1}{n} - \frac{1}{n^6}\right)} = \frac{1 + \frac{1}{n^2} - \frac{1000}{n^6}}{7 + \frac{1}{n} - \frac{1}{n^6}}.$

Step 2: Take the limit as $n \to \infty$: As $n \to \infty$, terms with negative powers of $n$ approach 0: $\frac{1 + \frac{1}{n^2} - \frac{1000}{n^6}}{7 + \frac{1}{n} - \frac{1}{n^6}} \to \frac{1}{7}.$

Final Answers:

1. $\lim_{x \to 0} \frac{\cos x - 1 + \frac{1}{2}x^2}{x^4} = \frac{1}{24}$,
2. $\lim_{n \to \infty} \frac{n^6 + n^4 - 1000}{7n^6 + n^5 - 1} = \frac{1}{7}$.

Would you like me to explain any specific step in more detail or solve related problems?

Further Questions:

1. Can you derive the Taylor expansion for $\cos x$ up to higher orders?
2. How would you calculate the limit in (i) using L'Hôpital's Rule instead of Taylor expansion?
3. Can you evaluate a similar limit for a different function like $\sin x$?
4. For (ii), how does the behavior of $n^k$ as $n \to \infty$ affect such limits in general?
5. Can you find other ways to simplify large degree polynomial limits?

Tip:

When evaluating limits of rational functions as $n \to \infty$, always divide through by the highest power of $n$ in the denominator.