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I can help you solve the limit problems you uploaded. Let's go through each of them one by one:

### 1. \(\lim_{x \to \frac{\pi}{2}} \left( \sin\frac{1}{2}x + \cos x \right)\)
For this limit, simply substitute \(x = \frac{\pi}{2}\) directly:

\[
\sin\frac{1}{2}x = \sin\frac{1}{2}\left(\frac{\pi}{2}\right) = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}
\]
\[
\cos x = \cos\left(\frac{\pi}{2}\right) = 0
\]

Thus, the limit becomes:
\[
\lim_{x \to \frac{\pi}{2}} \left( \sin\frac{1}{2}x + \cos x \right) = \frac{\sqrt{2}}{2} + 0 = \frac{\sqrt{2}}{2}
\]

### 2. \(\lim_{x \to 0} \frac{4 - 4\cos^2 x}{x \sin 2x}\)
For this problem, let's simplify the expression:

We know the identity \(\cos^2 x = \frac{1 + \cos(2x)}{2}\), so:
\[
4 - 4\cos^2 x = 4 - 4\left(\frac{1 + \cos(2x)}{2}\right) = 4 - 2 - 2\cos(2x) = 2 - 2\cos(2x)
\]

Thus, the limit becomes:
\[
\lim_{x \to 0} \frac{2 - 2\cos(2x)}{x \sin 2x}
\]

Using the small-angle approximation for \(\cos(2x)\) and \(\sin(2x)\):
\[
\cos(2x) \approx 1 - \frac{(2x)^2}{2} \quad \text{and} \quad \sin(2x) \approx 2x \quad \text{as} \quad x \to 0
\]

Now, substitute these approximations:
\[
\frac{2 - 2(1 - 2x^2)}{x \cdot 2x} = \frac{2(2x^2)}{2x^2} = 2
\]

So, the limit is:
\[
\boxed{2}
\]

### 3. \(\lim_{x \to \infty} \left( \frac{\sin^2\frac{2}{x}}{x} + 2\sin\frac{1}{x} + 8 \right)\)
For this limit as \(x \to \infty\), observe the behavior of each term:

- \(\sin\left(\frac{2}{x}\right)\) and \(\sin\left(\frac{1}{x}\right)\) approach \(0\) as \(x \to \infty\), since the sine function of a small number approaches the small number itself.

Thus:
\[
\frac{\sin^2\frac{2}{x}}{x} \approx \frac{\left(\frac{2}{x}\right)^2}{x} = \frac{4}{x^3} \to 0
\]
\[
2\sin\frac{1}{x} \approx 2\left(\frac{1}{x}\right) = \frac{2}{x} \to 0
\]

Therefore, the limit becomes:
\[
\lim_{x \to \infty} \left( 0 + 0 + 8 \right) = 8
\]

So, the final answer is:
\[
\boxed{8}
\]

---

Do you want more details or have any further questions? Here are 5 related questions that can expand your understanding:

1. How do we use the small-angle approximation in limit problems?
2. What is the importance of direct substitution in limit evaluation?
3. How do trigonometric limits behave as the argument approaches zero or infinity?
4. How do L'Hopital's Rule and small-angle approximations help in solving indeterminate forms?
5. What are other common limit evaluation techniques besides substitution and L'Hopital’s Rule?

**Tip:** Always simplify trigonometric identities before applying limits, as it makes evaluating the limit much easier.

Tentukan nilai limit dari:
1. lim(x → π/2) (sin(1/2)x + cos x)
2. lim(x → 0) (4 - 4cos^2(x)) / (xsin(2x))
3. lim(x → ∞) (sin^2(2/x)/x + 2sin(1/x) + 8)

Learn how to solve trigonometric limit problems, including evaluating limits at π/2, 0, and infinity, using key concepts like L'Hopital's Rule and small-angle approximations.

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