Math Problem Statement

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)

Solution

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. limx1x1x1\lim_{x \to 1} \frac{x - 1}{\sqrt{x} - \sqrt{1}}

    Solution:

    \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. limx3(1x13)x3\lim_{x \to 3} \frac{\left(\frac{1}{x} - \frac{1}{3}\right)}{x - 3}

    Solution: Simplify: 3x3xx3=(x3)3x(x3)=13x\frac{\frac{3 - x}{3x}}{x - 3} = \frac{-(x - 3)}{3x(x - 3)} = \frac{-1}{3x} limx313x=19\lim_{x \to 3} \frac{-1}{3x} = \frac{-1}{9}

  3. limx01cosxx2\lim_{x \to 0} \frac{1 - \cos{x}}{x^2}

    Solution: Using the approximation 1cosxx221 - \cos{x} \approx \frac{x^2}{2} as x0x \to 0: limx01cosxx2=limx0x22x2=12\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. limx0xtanx1cos2x\lim_{x \to 0} \frac{x \tan{x}}{1 - \cos{2x}}

    Solution: Approximation for small xx: tanxx\tan{x} \approx x and 1cos2x2x21 - \cos{2x} \approx 2x^2: limx0xx2x2=limx0x22x2=12\lim_{x \to 0} \frac{x \cdot x}{2x^2} = \lim_{x \to 0} \frac{x^2}{2x^2} = \frac{1}{2}

  5. limx2x2x2+1\lim_{x \to \infty} \frac{2x^2}{x^2 + 1}

    Solution: Divide numerator and denominator by x2x^2: limx2x2x2+1=limx21+1x2=2\lim_{x \to \infty} \frac{2x^2}{x^2 + 1} = \lim_{x \to \infty} \frac{2}{1 + \frac{1}{x^2}} = 2

  6. limx2xxx+3x3/2\lim_{x \to \infty} \frac{2x \sqrt{x} - x + 3}{x^{3/2}}

    Solution: Factor out x3/2x^{3/2} from the numerator: limxx3/2(21x+3x3/2)x3/2=limx(20+0)=2\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. limxx22x32x2+4x+7\lim_{x \to \infty} \frac{x^2 - 2x - 3}{2x^2 + 4x + 7}

    Solution: Divide numerator and denominator by x2x^2: limxx22x32x2+4x+7=limx12x3x22+4x+7x2=12\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 xx 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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Limits
Algebraic Limits
Infinity Limits
L'Hopital's Rule
Rationalization

Formulas

lim x->a f(x)/g(x)
lim x->0 (1 - cos(x)) / x^2 = 1/2
lim x->infinity 2x^2 / (x^2 + 1)
lim x->infinity (ax + b) / (cx + d) = a/c

Theorems

L'Hopital's Rule
Squeeze Theorem
Limit of Polynomial Functions

Suitable Grade Level

Grades 11-12 or First Year University