Let's solve the two limit problems from the image.

Problem 8:

We are asked to find: $\lim_{x \to \frac{\pi}{2}} \frac{\sin\left(x - \frac{\pi}{2}\right)}{\sqrt{\frac{x}{2}} - \sqrt{\frac{\pi}{4}}}$

First, let's simplify the expression.

• $\sin(x - \frac{\pi}{2}) = -\cos(x)$ using the identity $\sin(x - \frac{\pi}{2}) = -\cos(x)$.
• As $x \to \frac{\pi}{2}$, $\cos(x) \to 0$, so the numerator tends to 0.
• In the denominator, as $x \to \frac{\pi}{2}$:
  • $\sqrt{\frac{x}{2}} \to \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2}$.
  • Hence, $\sqrt{\frac{x}{2}} - \sqrt{\frac{\pi}{4}} \to 0$.

Both the numerator and denominator tend to 0, so we can apply L'Hopital's Rule.

Now, differentiate the numerator and denominator with respect to $x$:

1. The derivative of the numerator, $\sin(x - \frac{\pi}{2})$, is $-\cos(x)$, and $\frac{d}{dx}(-\cos(x)) = \sin(x)$.
2. The derivative of the denominator, $\sqrt{\frac{x}{2}} - \sqrt{\frac{\pi}{4}}$, is: $\frac{d}{dx}\left(\sqrt{\frac{x}{2}}\right) = \frac{1}{2\sqrt{2x}} = \frac{1}{2\sqrt{x/2}}.$

Thus, applying L'Hopital's Rule, we get: $\lim_{x \to \frac{\pi}{2}} \frac{\sin(x)}{\frac{1}{2\sqrt{x/2}}}.$ Now, substitute $x = \frac{\pi}{2}$ into this expression:

• $\sin\left(\frac{\pi}{2}\right) = 1$,
• $\frac{1}{2\sqrt{\frac{\pi}{4}}} = \frac{1}{2\frac{\sqrt{\pi}}{2}} = \frac{1}{\sqrt{\pi}}$.

Hence, the limit becomes: $\frac{1}{\frac{1}{\sqrt{\pi}}} = \sqrt{\pi}.$

So, the answer is C. $\sqrt{\pi}$.

Problem 9:

We are asked to find: $\lim_{x \to 3} \frac{(x + 2)\tan(x - 3)}{2x^2 - 5x - 3}.$

First, substitute $x = 3$:

• The numerator becomes $5 \cdot \tan(0) = 5 \cdot 0 = 0$.
• The denominator becomes $2(3)^2 - 5(3) - 3 = 18 - 15 - 3 = 0$.

Since both the numerator and denominator tend to 0, we can apply L'Hopital's Rule.

Differentiate the numerator and denominator:

1. The derivative of $(x + 2)\tan(x - 3)$ using the product rule is: $\frac{d}{dx}((x + 2)\tan(x - 3)) = (1)(\tan(x - 3)) + (x + 2)\sec^2(x - 3).$ At $x = 3$, $\tan(0) = 0$ and $\sec^2(0) = 1$, so this becomes: $0 + (5)(1) = 5.$

2. The derivative of the denominator, $2x^2 - 5x - 3$, is: $\frac{d}{dx}(2x^2 - 5x - 3) = 4x - 5.$ At $x = 3$, this becomes $4(3) - 5 = 12 - 5 = 7$.

Now, applying L'Hopital's Rule, we get: $\frac{5}{7}.$

So, the answer is C. $\frac{5}{7}$.

Let me know if you want further details or clarifications. Here are 5 related questions you might find interesting:

1. How do you determine when to apply L'Hopital's Rule?
2. What is the geometric interpretation of limits involving trigonometric functions?
3. How can you apply Taylor series to simplify limit problems?
4. What are the conditions for applying L'Hopital’s Rule?
5. Can you solve limits using factorization instead of L'Hopital's Rule?

Tip: Always check if direct substitution works before using advanced techniques like L'Hopital’s Rule.