Math Problem Statement

What is the value of the limit: lim(x -> -2-) ((2x^2 + 7x - 4) / (x^2 + x - 2))?

Solution

The limit you provided is:

limx22x2+7x4x2+x2\lim_{x \to -2^-} \frac{2x^2 + 7x - 4}{x^2 + x - 2}

To solve this, let's follow these steps:

1. Factor the numerator and denominator

We will start by factoring both the numerator and denominator, if possible.

Numerator:

2x2+7x42x^2 + 7x - 4

To factor, we need two numbers that multiply to 2×(4)=82 \times (-4) = -8 and add to 77. The factors are 88 and 1-1. So we can rewrite the middle term:

2x2+8xx4=2x(x+4)1(x+4)=(2x1)(x+4)2x^2 + 8x - x - 4 = 2x(x + 4) - 1(x + 4) = (2x - 1)(x + 4)

Denominator:

x2+x2x^2 + x - 2

We need two numbers that multiply to 2-2 and add to 11. The factors are 22 and 1-1:

x2+x2=(x1)(x+2)x^2 + x - 2 = (x - 1)(x + 2)

2. Rewrite the expression with factored terms

Now we have:

(2x1)(x+4)(x1)(x+2)\frac{(2x - 1)(x + 4)}{(x - 1)(x + 2)}

3. Evaluate the limit as x2x \to -2^-

Substitute x=2x = -2 into the factored expression:

(2(2)1)((2)+4)((2)1)((2)+2)\frac{(2(-2) - 1)((-2) + 4)}{((-2) - 1)((-2) + 2)}

This simplifies to:

(41)(2+4)(21)(2+2)=(5)(2)(3)(0)\frac{(-4 - 1)(-2 + 4)}{(-2 - 1)(-2 + 2)} = \frac{(-5)(2)}{(-3)(0)}

The denominator becomes 00, which means the function is undefined at x=2x = -2, and we need to analyze the behavior of the function as x2x \to -2^- (from the left).

  • As x2x \to -2^-, x+2x + 2 approaches 00^- (a small negative value).
  • The numerator will still approach (5)(2)=10(-5)(2) = -10.
  • The denominator will approach a small negative value because x1x - 1 is negative and x+2x + 2 approaches 00^-.

Thus, as x2x \to -2^-, the expression behaves like:

10small negative number+\frac{-10}{\text{small negative number}} \to +\infty

Conclusion:

The limit as x2x \to -2^- of the given function is ++\infty.

Would you like further details or clarifications?


Follow-up Questions:

  1. How do one-sided limits differ from two-sided limits?
  2. Why does the denominator becoming zero affect the limit?
  3. How can we use L'Hopital's Rule in limit problems involving indeterminate forms?
  4. Can the function in the example be continuous at x=2x = -2?
  5. What are the implications if we approach x=2x = -2 from the right side?

Tip:

Always check if factoring helps simplify a limit expression before applying advanced techniques like L'Hopital's Rule!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Limits
Factoring Quadratics
One-sided limits

Formulas

Factoring quadratics: ax^2 + bx + c = 0
Limit of a rational function

Theorems

Limits and continuity
Factoring to simplify expressions

Suitable Grade Level

Grades 10-12