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Let's analyze the limits provided in the problem one by one. The key hint for solving these is to check the behavior of the denominator first. Here's a breakdown:

### Step-by-Step Strategy:
1. **Check the denominator**: If it doesn't go to 0 as \( t \to \frac{\pi}{2} \), evaluate the limit directly.
2. If **denominator goes to 0** and numerator doesn't, the limit may tend towards infinity or negative infinity.
3. If **both go to 0**: Use algebraic techniques like rationalizing, or apply L'Hôpital's Rule.

### Given Limits:

For all limits in this problem, the denominator is \( 2 \cos 2t \) or a similar form. At \( t = \frac{\pi}{2} \), \( \cos 2t = \cos(\pi) = -1 \), so the denominator doesn't approach 0. This simplifies many limits significantly. Let’s go case by case:

### Limit (a):
\[
\lim_{t \to \frac{\pi}{2}} \frac{\sin t - \sqrt{\sin^2 t + 6 \cos 2 t}}{2 \cos 2 t}
\]
At \( t = \frac{\pi}{2} \), \( \sin \frac{\pi}{2} = 1 \) and \( \cos \pi = -1 \).
- Denominator: \( 2 \cos 2t = 2 \times (-1) = -2 \).
- Numerator: \( \sin \frac{\pi}{2} = 1 \) and \( \cos \pi = -1 \), so \( \sqrt{\sin^2(\frac{\pi}{2}) + 6 \cos 2t} = \sqrt{1 + 6 \times (-1)} = \sqrt{1 - 6} = \sqrt{-5} \), which is **undefined** in the real number system.

Thus, **limit (a) does not exist** (DNE) due to the square root of a negative number.

### Limit (b):
\[
\lim_{t \to \frac{\pi}{2}} \frac{\sin t + \sqrt{\sin^2 t + 6 \cos 2 t}}{2 \cos 2 t}
\]
The denominator behaves similarly: \( 2 \cos 2t = -2 \). For the numerator:
- The square root expression is still \( \sqrt{-5} \), which is undefined.
Thus, **limit (b) also does not exist (DNE)**.

### Limit (c):
\[
\lim_{t \to \frac{\pi}{2}} \frac{\cos t - \sqrt{\sin^2 t + 6 \cos 2 t}}{2 \cos 2 t}
\]
Here, \( \cos \frac{\pi}{2} = 0 \) and the same issue arises with the square root of a negative number.
Thus, **limit (c) does not exist (DNE)**.

### Limit (d):
\[
\lim_{t \to \frac{\pi}{2}} \frac{\cos t + \sqrt{\sin^2 t + 6 \cos 2 t}}{2 \cos 2 t}
\]
- The square root term is still \( \sqrt{-5} \), leading to **DNE**.

### Limit (e), (f), and (g):
These involve a similar pattern with \( 2 \sin 2t \) in the denominator. The square root \( \sqrt{-5} \) in each case still leads to **DNE** for all of these limits.

### Conclusion:
For **all limits (a) through (g)**, the result is **DNE (Does Not Exist)** because of the appearance of the square root of a negative number under the given conditions.

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Would you like more details or further clarification on any of these limits? Here are some questions you could explore:

1. What is the importance of checking the denominator first in limits involving quotients?
2. How would L'Hôpital’s rule be applied in cases where both the numerator and denominator approach zero?
3. Can limits involving square roots of negative numbers ever exist in the real number system?
4. How does the behavior of trigonometric functions like \( \sin t \) and \( \cos t \) affect the limits as \( t \to \frac{\pi}{2} \)?
5. What are some alternative techniques to solve indeterminate forms in limit problems?

**Tip**: Always check whether the numerator and denominator both tend to zero or infinity before applying L'Hôpital's Rule or rationalizing.

Calculate the limits for the given expressions. If the limit is ∞ or -∞, write infinity or -infinity. If the limit does not exist, write DNE.

Explore the detailed step-by-step process for calculating limits involving trigonometric functions and square roots. Learn the use of L'Hopital's Rule and the analysis of indeterminate forms in this calculus-level problem.

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