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The image contains a list of seven limit problems in Indonesian. I will go through each limit problem one by one and provide their solutions.

1. \(\lim_{x \to 1} \frac{x - 1}{\sqrt{x} - \sqrt{1}}\)

   **Solution**: 
   \[
   \frac{0}{0} \text{ form, so rationalize the denominator: }
   \frac{x - 1}{\sqrt{x} - 1} \times \frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \frac{(x - 1)(\sqrt{x} + 1)}{x - 1} = \sqrt{x} + 1
   \]
   \[
   \lim_{x \to 1} (\sqrt{x} + 1) = \sqrt{1} + 1 = 2
   \]

2. \(\lim_{x \to 3} \frac{\left(\frac{1}{x} - \frac{1}{3}\right)}{x - 3}\)

   **Solution**: 
   Simplify:
   \[
   \frac{\frac{3 - x}{3x}}{x - 3} = \frac{-(x - 3)}{3x(x - 3)} = \frac{-1}{3x}
   \]
   \[
   \lim_{x \to 3} \frac{-1}{3x} = \frac{-1}{9}
   \]

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

   **Solution**:
   Using the approximation \(1 - \cos{x} \approx \frac{x^2}{2}\) as \(x \to 0\):
   \[
   \lim_{x \to 0} \frac{1 - \cos{x}}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2}}{x^2} = \frac{1}{2}
   \]

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

   **Solution**: 
   Approximation for small \(x\): \(\tan{x} \approx x\) and \(1 - \cos{2x} \approx 2x^2\):
   \[
   \lim_{x \to 0} \frac{x \cdot x}{2x^2} = \lim_{x \to 0} \frac{x^2}{2x^2} = \frac{1}{2}
   \]

5. \(\lim_{x \to \infty} \frac{2x^2}{x^2 + 1}\)

   **Solution**:
   Divide numerator and denominator by \(x^2\):
   \[
   \lim_{x \to \infty} \frac{2x^2}{x^2 + 1} = \lim_{x \to \infty} \frac{2}{1 + \frac{1}{x^2}} = 2
   \]

6. \(\lim_{x \to \infty} \frac{2x \sqrt{x} - x + 3}{x^{3/2}}\)

   **Solution**: 
   Factor out \(x^{3/2}\) from the numerator:
   \[
   \lim_{x \to \infty} \frac{x^{3/2}(2 - \frac{1}{\sqrt{x}} + \frac{3}{x^{3/2}})}{x^{3/2}} = \lim_{x \to \infty} (2 - 0 + 0) = 2
   \]

7. \(\lim_{x \to \infty} \frac{x^2 - 2x - 3}{2x^2 + 4x + 7}\)

   **Solution**:
   Divide numerator and denominator by \(x^2\):
   \[
   \lim_{x \to \infty} \frac{x^2 - 2x - 3}{2x^2 + 4x + 7} = \lim_{x \to \infty} \frac{1 - \frac{2}{x} - \frac{3}{x^2}}{2 + \frac{4}{x} + \frac{7}{x^2}} = \frac{1}{2}
   \]

Would you like further details on any of these solutions?

### Related Questions:
1. How can we apply L'Hopital's Rule to solve indeterminate limits?
2. What are some common limit properties that simplify limit calculations?
3. How do approximations of trigonometric functions help in limit evaluations?
4. What is the significance of dividing by the highest power of \(x\) in the denominator?
5. How do we handle limits approaching infinity in rational functions?

**Tip**: Rationalizing the denominator is a powerful technique to simplify limits involving square roots.

Tentukan nilai limit dari soal berikut ini: lim x->1 (x - 1) / (sqrt(x) - sqrt(1)), lim x->3 [(1/x) - (1/3)] / (x - 3), lim x->0 (1 - cos(x)) / x^2, lim x->0 (x tan(x)) / (1 - cos(2x)), lim x->infinity (2x^2) / (x^2 + 1), lim x->infinity [(2x sqrt(x) - x + 3) / x^(3/2)], lim x->infinity (x^2 - 2x - 3) / (2x^2 + 4x + 7)

Solve advanced limit problems involving trigonometric functions, rational expressions, and limits approaching infinity. Techniques include rationalization and L'Hopital's Rule. Suitable for high school and first-year university students.

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