Math Problem Statement

Use the ε-δ definition of the limit for functions to show lim (x → 1) (x^2 - 3x) = -2.

Solution

To prove limx1(x23x)=2\lim_{x \to 1} (x^2 - 3x) = -2 using the ε\varepsilon-δ\delta definition of the limit, we need to follow these steps:


ε\varepsilon-δ\delta Definition

The statement limx1f(x)=2\lim_{x \to 1} f(x) = -2 means: For every ε>0\varepsilon > 0, there exists a δ>0\delta > 0 such that whenever 0<x1<δ0 < |x - 1| < \delta, it follows that f(x)(2)<ε|f(x) - (-2)| < \varepsilon.

Here f(x)=x23xf(x) = x^2 - 3x.


Proof Steps

  1. Start with the inequality f(x)(2)<ε|f(x) - (-2)| < \varepsilon: f(x)(2)=(x23x)+2=x23x+2.|f(x) - (-2)| = |(x^2 - 3x) + 2| = |x^2 - 3x + 2|.

  2. Factorize the expression x23x+2x^2 - 3x + 2: x23x+2=(x1)(x2).x^2 - 3x + 2 = (x - 1)(x - 2). Therefore, x23x+2=(x1)(x2).|x^2 - 3x + 2| = |(x - 1)(x - 2)|.

  3. Simplify the inequality: (x1)(x2)<ε.|(x - 1)(x - 2)| < \varepsilon.

  4. Control the term x2|x - 2|: When xx is close to 11, say x1<1|x - 1| < 1, it follows that: x2=(x1)1x1+1<1+1=2.|x - 2| = |(x - 1) - 1| \leq |x - 1| + |1| < 1 + 1 = 2.

    So, x2<2|x - 2| < 2 if x1<1|x - 1| < 1.

  5. Bound (x1)(x2)|(x - 1)(x - 2)|: Using x2<2|x - 2| < 2, we can write: (x1)(x2)=x1x2<x12.|(x - 1)(x - 2)| = |x - 1| \cdot |x - 2| < |x - 1| \cdot 2.

  6. Choose δ\delta: To ensure (x1)(x2)<ε|(x - 1)(x - 2)| < \varepsilon, we require: x12<ε    x1<ε2.|x - 1| \cdot 2 < \varepsilon \implies |x - 1| < \frac{\varepsilon}{2}. Therefore, choose δ=min(1,ε2)\delta = \min\left(1, \frac{\varepsilon}{2}\right).

  7. Verify: If 0<x1<δ0 < |x - 1| < \delta, then: (x1)(x2)<x12<ε22=ε.|(x - 1)(x - 2)| < |x - 1| \cdot 2 < \frac{\varepsilon}{2} \cdot 2 = \varepsilon. Thus, f(x)(2)<ε|f(x) - (-2)| < \varepsilon, completing the proof.


Conclusion

We have shown that for any ε>0\varepsilon > 0, choosing δ=min(1,ε2)\delta = \min\left(1, \frac{\varepsilon}{2}\right) satisfies the condition of the ε\varepsilon-δ\delta definition of the limit. Hence, limx1(x23x)=2\lim_{x \to 1} (x^2 - 3x) = -2.


Do you want further clarification on any step or more examples? 😊


Related Questions

  1. How can you interpret the ε\varepsilon-δ\delta definition geometrically?
  2. What changes if the limit were xcx \to c where c1c \neq 1?
  3. How would you prove a one-sided limit using the ε\varepsilon-δ\delta definition?
  4. What happens if f(x)f(x) is not continuous at the point of the limit?
  5. Can you use the ε\varepsilon-δ\delta definition to prove the limit of a rational function?

Tip

Always start by simplifying the function and bounding terms before choosing δ\delta. This helps streamline your proof!

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Math Problem Analysis

Mathematical Concepts

Limits
Algebraic Manipulation
ε-δ Definition of Limits

Formulas

|f(x) - L| < ε for 0 < |x - c| < δ

Theorems

ε-δ Definition of Limits

Suitable Grade Level

Undergraduate Calculus (Introductory Level)