Problem 34 - Calculus: Early Transcendentals

Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x-3}$$

Step-by-Step Solution

Answer

Question: Find the limit of the function \(\lim_{x \rightarrow 3} \frac{x^2-2x-3}{x-3}\). Answer: The limit of the given function as x approaches 3 is 4.

Step 1: Identify the problem at the limit point

At x=3, the denominator, x-3, becomes zero. To proceed to finding the limit, let's simplify the given function and see if the zero at the denominator can be removed.

Step 2: Factor the numerator

To simplify the given function, we can factor the numerator and see if there is a common factor with the denominator. The numerator is a quadratic expression, factor it as follows: \(x^2 - 2x -3 = (x-3)(x+1)\) So the function becomes: \(\frac{x^{2}-2 x-3}{x-3} = \frac{(x-3)(x+1)}{x-3}\)

Step 3: Cancel the common terms

Notice that both the numerator and the denominator have a common factor of (x-3). Cancel it out: \(\frac{(x-3)(x+1)}{x-3} = \frac{\cancel{(x-3)}(x+1)}{\cancel{x-3}} = x+1\) The simplified function is \(x + 1\).

Step 4: Evaluate the limit

Now that we have simplified the function, let's evaluate the limit: \(\lim_{x \rightarrow 3} (x+1) = (3) + 1\)

Step 5: Write the final answer

Calculating the limit gives us: \(\lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x-3} = (3) + 1 = 4\)

Key Concepts

Simplifying Rational Expressions

Simplifying rational expressions is an essential step in solving limits problems, especially when dealing with expressions that have a denominator potentially reaching zero. In our exercise, the expression is \( \frac{x^{2}-2x-3}{x-3} \). At \( x=3 \), the denominator becomes zero, which could lead to an indeterminate form that's undefined in standard mathematics. To address this issue, we simplify the rational expression.The simplification process involves finding a common factor in both the numerator and denominator, which can then be canceled, removing the potential for division by zero. In our example, we noticed that the expression \( x^2 - 2x - 3 \) can be factored to \((x-3)(x+1)\). This means we can rewrite the original problematic expression as:\[ \frac{(x-3)(x+1)}{x-3} \]Now, by canceling the common factor \((x-3)\), we are left with \(x+1\), a simplified expression without terms dividing by zero. Simplifying rational expressions is valuable as it often reveals the true behavior of the function, allowing for efficient limit evaluation.

Factoring Quadratics

Factoring quadratics is a common approach when simplifying expressions, especially if dealing with polynomial numerators. In this example, the quadratic expression \( x^2 - 2x - 3 \) was given. To factor it, we need to find two numbers that multiply to \(-3\) (the constant term) and add to \(-2\) (the coefficient of the middle term, \(-2x\)).Here’s how we break it down:

• First, recognize \( -3 \) can be written as \(-3 = 1 \times (-3)\) or \(-1 \times 3\).
• Only the pair \(1\) and \(-3\) sums to \(-2\).

With these numbers, we factor the quadratic as \((x-3)(x+1)\). Factoring is a crucial skill not just for limit problems but for many algebraic manipulations. Understanding how quadratic expressions can be broken down helps in revealing simplifications that may not be obvious at first glance.

Limit Evaluation

Limit evaluation is a fundamental technique used in calculus to find the value a function approaches as the input approaches a certain point. In this problem, we aimed to evaluate the limit \( \lim_{x \rightarrow 3} \frac{x^{2}-2x-3}{x-3} \). Initially, direct substitution into this expression would have led to division by zero, prompting us to simplify the expression first.After simplifying to \( x+1 \), limit evaluation becomes straightforward. We now substitute \( x=3 \) directly into the simplified expression:\[ \lim_{x \rightarrow 3} (x+1) = 3 + 1 = 4 \]This process demonstrates that simplifying expressions often reveals a function's true behavior near the limit point. Proper evaluation of limits is essential for understanding continuity, derivatives, and the overall behavior of functions. Learning to navigate indeterminate forms by simplification and strategic factoring is key for anyone studying calculus.