Calculus Intro

Limit

Concept of Approaching, Indeterminate Forms, and Algebraic Evaluation.

1. Concept of Limit (\(x \to a\))

The expression \(\lim_{x \to a} f(x) = L\) means that as the variable \(x\) gets arbitrarily close to \(a\) (but \(x \neq a\)), the value of the function \(f(x)\) approaches \(L\).

Term Meaning
Left-Hand Limit (LHL) \(\lim_{x \to a^-} f(x)\) (Approaching \(a\) from values less than \(a\))
Right-Hand Limit (RHL) \(\lim_{x \to a^+} f(x)\) (Approaching \(a\) from values greater than \(a\))
Existence of Limit The limit exists if and only if: LHL = RHL = Finite Value.

2. Direct Substitution & Indeterminate Forms

The first step in evaluating any algebraic limit is direct substitution. If substitution yields a real number, that is the limit. If it yields an undefined form, further algebraic manipulation is required.

• Determinate Form: \(\lim_{x \to 2} (x^2 + 3) = 2^2 + 3 = 7\)

• Indeterminate Forms: \(\frac{0}{0}, \frac{\infty}{\infty}, \infty - \infty, 0 \times \infty\)

When \(\frac{0}{0}\) occurs, it implies there is a common factor in both the numerator and the denominator causing the expression to become zero at \(x = a\).

3. Factorization Method (Variable Cancellation)

When a rational function evaluates to \(\frac{0}{0}\), we factorize the numerator and denominator to isolate and cancel the problematic variable terms. Crucially, we can only cancel these variables because \(x \to a\) strictly means \(x \neq a\), ensuring the denominator is not actually zero during the cancellation process.

Example: Evaluating a \(\frac{0}{0}\) Form

Evaluate: \(\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\)

Step Action / Logic Mathematical Expression
1. Test Substitution Substitute \(x=3\) directly to check form. \(\frac{3^2 - 9}{3 - 3} = \frac{0}{0}\) (Indeterminate)
2. Factorize Apply difference of squares to the numerator. \(\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}\)
3. Cancel Common Factors Cancel \((x - 3)\) from numerator and denominator. Since \(x \to 3\), \(x - 3 \neq 0\), making cancellation mathematically valid. \(\lim_{x \to 3} (x + 3)\)
4. Final Substitution Substitute \(x=3\) into the simplified expression. \(3 + 3 = \mathbf{6}\)

4. Rationalization Method

Used primarily when the function involves square roots and yields a \(\frac{0}{0}\) form upon direct substitution. We multiply the numerator and denominator by the conjugate of the irrational term.

Example: Rationalizing the Numerator

Evaluate: \(\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}\)

Step Action Result
1. Test Substitute \(x=0\) \(\frac{\sqrt{4} - 2}{0} = \frac{0}{0}\)
2. Multiply by Conjugate Multiply by \(\frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}\) \(\lim_{x \to 0} \frac{(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2)}{x(\sqrt{x + 4} + 2)}\)
3. Simplify Numerator Apply \((a-b)(a+b) = a^2 - b^2\) \(\lim_{x \to 0} \frac{(x + 4) - 4}{x(\sqrt{x + 4} + 2)}\) = \(\lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)}\)
4. Cancel and Substitute Cancel \(x\) (valid since \(x \neq 0\)), then substitute \(x=0\) \(\frac{1}{\sqrt{0 + 4} + 2} = \frac{1}{2 + 2} = \mathbf{\frac{1}{4}}\)