Math Problem Statement

Evaluate the following three limits: 1. \( \lim_{x \to 0^-} \frac{3x + 4}{x^2} \) 2. \( \lim_{x \to +\infty} \frac{3 - 2x^2}{x^2 + 1} \) 3. \( \lim_{x \to -\infty} \frac{x + 1}{x^3 - 1} \)

Solution

Let's evaluate the three limits one by one. Here are the step-by-step solutions:


1. Limit 1: limx03x+4x2\lim_{x \to 0^-} \frac{3x + 4}{x^2}

  • For x0x \to 0^-, the denominator x2x^2 remains positive because it is squared, but the numerator depends on xx.
  • When x0x \to 0^-, 3x3x becomes slightly negative, and 44 remains positive.
  • Therefore, 3x+443x + 4 \to 4 (slightly less than 4 from the left).

So,
limx03x+4x2=small negativesmall positive.\lim_{x \to 0^-} \frac{3x + 4}{x^2} = \frac{\text{small negative}}{\text{small positive}} \to -\infty.

Conclusion:
limx03x+4x2=.\lim_{x \to 0^-} \frac{3x + 4}{x^2} = -\infty.


2. Limit 2: limx+32x2x2+1\lim_{x \to +\infty} \frac{3 - 2x^2}{x^2 + 1}

  • The dominant term in both the numerator and denominator is x2x^2 as x+x \to +\infty.
  • Divide both numerator and denominator by x2x^2 to simplify:

32x2x2+1=3x221+1x2.\frac{3 - 2x^2}{x^2 + 1} = \frac{\frac{3}{x^2} - 2}{1 + \frac{1}{x^2}}.

  • As x+x \to +\infty, terms involving 1x20\frac{1}{x^2} \to 0, so the expression simplifies to:

021+0=2.\frac{0 - 2}{1 + 0} = -2.

Conclusion:
limx+32x2x2+1=2.\lim_{x \to +\infty} \frac{3 - 2x^2}{x^2 + 1} = -2.


3. Limit 3: limxx+1x31\lim_{x \to -\infty} \frac{x + 1}{x^3 - 1}

  • For xx \to -\infty, the dominant term in the numerator is xx, and in the denominator, x3x^3 dominates.
  • Divide both numerator and denominator by x3x^3:

x+1x31=xx3+1x311x3.\frac{x + 1}{x^3 - 1} = \frac{\frac{x}{x^3} + \frac{1}{x^3}}{1 - \frac{1}{x^3}}.

Simplify each term:

  • xx3=1x20\frac{x}{x^3} = \frac{1}{x^2} \to 0 as xx \to -\infty.
  • 1x30\frac{1}{x^3} \to 0 as xx \to -\infty.

Thus, the limit simplifies to:

0+010=0.\frac{0 + 0}{1 - 0} = 0.

Conclusion:
limxx+1x31=0.\lim_{x \to -\infty} \frac{x + 1}{x^3 - 1} = 0.


Final Answers:

  1. limx03x+4x2=\lim_{x \to 0^-} \frac{3x + 4}{x^2} = -\infty
  2. limx+32x2x2+1=2\lim_{x \to +\infty} \frac{3 - 2x^2}{x^2 + 1} = -2
  3. limxx+1x31=0\lim_{x \to -\infty} \frac{x + 1}{x^3 - 1} = 0

Do you need further details or have any additional questions?

Here are 5 related questions for practice:

  1. Evaluate limx5x3x2x3+3x\lim_{x \to \infty} \frac{5x^3 - x}{2x^3 + 3x}.
  2. Find limx0+2x+1x2\lim_{x \to 0^+} \frac{2x + 1}{x^2}.
  3. Calculate limxx23xx2+2\lim_{x \to -\infty} \frac{x^2 - 3x}{x^2 + 2}.
  4. Evaluate limx4x2x+5x2+7\lim_{x \to \infty} \frac{4x^2 - x + 5}{x^2 + 7}.
  5. Find limx0sinxx\lim_{x \to 0} \frac{\sin x}{x}.

Tip:

When evaluating limits involving large xx, always focus on the dominant terms (highest powers of xx) in the numerator and denominator.

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Math Problem Analysis

Mathematical Concepts

Limits
Dominant terms in polynomials
Behavior of functions as x approaches zero or infinity

Formulas

For large x, dominant terms determine the behavior of polynomials.
Limits at infinity: Divide numerator and denominator by the highest power of x.
For x approaching 0, consider sign and behavior of terms.

Theorems

Limits of rational functions
Behavior of polynomials at infinity

Suitable Grade Level

Undergraduate Calculus or AP Calculus (Grade 11-12)