Problem 21 - Calculus Early Transcendentals

Question

Prove the statement using the \(\varepsilon, \delta\) definition of limit. \(\lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x-2}=5\)

Step-by-Step Solution

Answer

For any \(\varepsilon > 0\), let \(\delta = \varepsilon\), then \(\left| \frac{x^{2}+x-6}{x-2} - 5 \right| < \varepsilon\) for \(0 < |x-2| < \delta\).

Step 1: Simplify the Expression

To prove the limit, first try simplifying the expression \(\frac{x^2+x-6}{x-2}\). Notice that the numerator \(x^2 + x - 6\) can be factored as \((x-2)(x+3)\). Simplifying the expression gives:\[\frac{(x-2)(x+3)}{x-2} = x+3\text{ for }x eq 2.\]

Step 2: Identify the Limit Value

Now that the expression simplifies to \(x + 3\), evaluate \(\lim _{x \rightarrow 2} (x + 3)\). Substituting \(x = 2\) into \(x + 3\), we find:\[2 + 3 = 5.\]Thus, the limit value is \(5\).

Step 3: Set Up the \(\varepsilon, \delta\) Definition

We need to show that for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that:\[0 < |x - 2| < \delta \implies \left| \frac{x^{2}+x-6}{x-2} - 5 \right| < \varepsilon.\]Since\(\frac{x^{2}+x-6}{x-2} = x+3\), this simplifies to:\[|x+3 - 5| = |x-2|.\]

Step 4: Choose \(\delta\) Based on \(\varepsilon\)

From the equation |x-2| < ε, we note that \(\delta\) can be directly \(\varepsilon\) since \(|x+3 - 5| = |x-2|\). To meet the definition, set \(\delta = \varepsilon\).

Step 5: Apply the \(\varepsilon, \delta\) Condition

With \(\delta = \varepsilon\), for any \(\varepsilon > 0\), choosing \(\delta = \varepsilon\), ensures \[0 < |x - 2| < \delta \implies |x - 2| < \varepsilon.\]Hence, for all \(x\) in the interval \(0 < |x - 2| < \delta\), \(\left| \frac{x^{2}+x-6}{x-2} - 5 \right| < \varepsilon\).This confirms the limit using the \(\varepsilon, \delta\) definition.

Key Concepts

Calculus Limits

Calculus limits are fundamental in understanding how functions behave as inputs approach certain values. The statement \(\lim_{x \rightarrow a} f(x) = L\) means that as \(x\) gets closer to \(a\), the function \(f(x)\) gets closer to \(L\). This concept is essential in analyzing the behavior of functions, especially when they are not well-defined or have discontinuities at particular points.

• Limits describe the behavior of functions near specific points.
• Understanding limits is crucial for derivatives and integrals in calculus.

In our example, we determine that as \(x\) approaches 2, the expression \(\frac{x^2 + x - 6}{x - 2}\) approaches the value of 5. Through the limit, we analyze the behavior of the expression as it nears the point of discontinuity, even though the division by zero occurs at \(x = 2\).

Simplifying Rational Expressions

A rational expression can often be simplified by factoring. This process involves finding and canceling common factors in the numerator and the denominator.

• Factoring helps in removing discontinuities in functions.
• Simplification aids in limit evaluation by eliminating problematic points like division by zero.

For our problem, \(\frac{x^2 + x - 6}{x - 2}\) is simplified by factoring the numerator as \((x-2)(x+3)\). This leads to the cancellation of the \(x-2\) term, leaving \(x+3\) for \(x eq 2\). This step turns a potentially troublesome problem into a simpler one, making limit evaluation straightforward. Remember, always look for factors that can cancel out before evaluating limits to simplify expressions effectively.

Limit Evaluation

Limit evaluation involves determining the value a function approaches as the input approaches a certain point. Once a rational expression is simplified, substitution becomes easier and more accurate.

• Evaluate by substituting the limit point into the simplified function.
• Ensure no division by zero or indeterminate forms remain.

After simplifying \(\frac{x^2 + x - 6}{x - 2}\) to \(x+3\), you can substitute \(x = 2\) directly into \(x + 3\), yielding \(5\). This illustrates how simple substitution, post-simplification, effectively determines the limit. In general, check the simplified expression to guarantee no undefined situations before substituting for the limit.

Delta-Epsilon Proof

The \(\varepsilon, \delta\) definition of a limit is a formal method of proving that a function approaches a specific limit. This definition ensures that a function gets arbitrarily close to a limit value within a certain distance from a given point.

• The selection of \(\delta\) depends on a given \(\varepsilon\).
• Proving a limit with \(\varepsilon, \delta\) involves showing \(|f(x) - L| < \varepsilon\) for \(0 < |x - a| < \delta\).

In our case, to prove \(\lim_{x \rightarrow 2} \frac{x^2 + x - 6}{x - 2} = 5\), we show that \(|x + 3 - 5| < \varepsilon\) simplifies to \(|x - 2| < \varepsilon\). Therefore, set \(\delta = \varepsilon\) ensures the definition is satisfied, proving the limit rigorously. This proof technique is invaluable for confirming behavior of functions at given points according to formal mathematical standards.