Evaluate the limit \( \lim_{x o 3} rac{x^2 - 9}{x - 3} \).
["# Evaluate the Limit:\n( \lim_{x \ o 3} \frac{x^2 - 9}{x - 3} )", "Calculating limits is a fundamental skill in calculus, especially when dealing with rational expressions that appear undefined at certain input values. One common challenge students face—particularly near ( x = 3 )—is evaluating the limit of ( \frac{x^2 - 9}{x - 3} ) as ( x ) approaches 3. In this article, we’ll explore how to properly evaluate this limit, analyze its meaning, and explain the broader significance of removable discontinuities in rational functions.", "## Understanding the Expression", "The expression\n[\n\frac{x^2 - 9}{x - 3}\n]\nis undefined at ( x = 3 ), because plugging in 3 gives ( \frac{0}{0} ), an indeterminate form. However, before concluding the limit does not exist, it’s essential to simplify the expression and investigate the behavior as ( x ) approaches 3 from both sides.", "## Step 1: Factor the Numerator", "Notice that the numerator ( x^2 - 9 ) is a difference of squares:", "[\nx^2 - 9 = (x - 3)(x + 3)\n]", "Rewriting the limit expression, we get:", "[\n\lim_{x \ o 3} \frac{(x - 3)(x + 3)}{x - 3}\n]", "For all ( x <br/>\ne 3 ), the ( (x - 3) ) terms cancel neatly, simplifying the expression to:", "[\n\lim_{x \ o 3} (x + 3)\n]", "## Step 2: Evaluate the Simplified Limit", "Now the limit simplifies to a continuous function:", "[\n\lim_{x \ o 3} (x + 3) = 3 + 3 = 6\n]", "Because this simplified expression is defined and continuous at ( x = 3 ), we can substitute directly:", "[\n\lim_{x \ o 3} \frac{x^2 - 9}{x - 3} = 6\n]", "## Why This Limit Exists Despite the Indeterminate Form", "The expression ( \frac{x^2 - 9}{x - 3} ) appears undefined at ( x = 3 ), but the key insight is that the original rational function has a removable discontinuity (a hole) at ( x = 3 ). The factor ( (x - 3) ) cancels, revealing a smooth, linear function that approaches 6 as ( x ) nears 3 from both the left and the right.", "L’Hôpital’s Rule could also confirm the result, but caution is needed since L’Hôpital applies only when the limit is of the form ( \frac{0}{0} ) or ( \pm\infty/\pm\infty ), and here direct cancellation suffices.", "## Practical Implications and Wider Insights", "Understanding how to simplify limits by factoring is crucial when analyzing real-world functions in physics, engineering, and economics, where localized behavior near critical points determines system behavior. Moreover, recognizing removable discontinuities allows for continuous function representation through domain-aware simplifications.", "This example underscores the importance of algebraic manipulation in calculus: even when a function seems undefined at a point, a careful look often reveals that the limit exists and can be computed directly through simplification.", "## Conclusion", "The limit evaluates cleanly using algebraic factoring:", "[\n\lim_{x \ o 3} \frac{x^2 - 9}{x - 3} = 6\n]", "This result confirms that although the original expression is undefined at ( x = 3 ), the function approaches 6 smoothly, illustrating a removable discontinuity and the power of simplification in evaluating real limits.", "Understanding and applying such techniques equips learners with robust tools to tackle more complex limits and deepen their mastery of calculus fundamentals.", "---", "Related Keywords: evaluate limit x to 3, limit of (x²−9)/(x−3), removable discontinuity, simplifying rational functions, calculus limit computation, 6 limit at 3, function continuity, algebra in calculus."]
