Q.57 The value of lim 𝑥2−9/𝑥2−4𝑥+3 is __________(rounded off to the nearest integer).
x–>3

The limit as x approaches 3 of (x² − 9)/(x² − 4x + 3) is a classic calculus problem that tests understanding of factoring, cancellation of common factors, and evaluating limits of rational functions. Solving this limit carefully avoids the 0/0 indeterminate form and shows how algebra simplifies limit evaluation .

Step-by-Step Solution

Consider \(\lim_{x \to 3} \frac{x^2 – 9}{x^2 – 4x + 3}\).

Direct Substitution

The expression becomes \(\frac{0}{0}\), an indeterminate form, so algebraic simplification is required .

Factor Numerator and Denominator

Numerator \(x^2 – 9\) is a difference of squares: \(\frac{x^2 – 9}{(x^2 – 4x + 3)} = \frac{(x-3)(x+3)}{(x^2 – 4x + 3)}\).

Denominator \(x^2 – 4x + 3\) factors as: \(x^2 – 4x + 3 = (x-3)(x-1)\).

Cancel Common Factor

For \(x \neq 3\), cancel the common factor \((x-3)\): \(\frac{(x-3)(x+3)}{(x-3)(x-1)} = \frac{x+3}{x-1}\).

The limit of the original function as \(x \to 3\) equals the limit of the simplified function \(\frac{x+3}{x-1}\) as \(x \to 3\) .

Evaluate Simplified Limit

Therefore, \(\lim_{x \to 3} \frac{x^2 – 9}{x^2 – 4x + 3} = 3\).

Conceptual Explanation

• Why factoring? Factoring exposes common factors causing the 0/0 form at x=3; cancelling removes the removable discontinuity.
• Why cancellation allowed? In limits, only behavior near the point matters, not the function value at x=3.
• Simplified function continuity: \(\frac{x+3}{x-1}\) is continuous at x=3 where denominator ≠ 0 .

Multiple-Choice Analysis

Common options: (A) 0 (B) 1 (C) 2 (D) 3

• 0 or 1: From direct substitution ignoring 0/0 form.
• 2: Algebraic error in factoring or substitution.
• 3 (Correct): Proper factoring, cancellation, substitution .

Key Takeaways