Problem 4 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow-1} \frac{x+1}{x^{2}-2 x-3} $$

Step-by-Step Solution

Answer

The limit is \(-\frac{1}{4}\).

Step 1: Substitute the Limit Value

Substitute \( x = -1 \) into the expression \( \frac{x+1}{x^{2}-2x-3} \) to see if it forms an indeterminate form. The substitution yields \( \frac{-1 + 1}{(-1)^2 - 2(-1) - 3} = \frac{0}{1 + 2 - 3} = \frac{0}{0}\), indicating an indeterminate form and suggesting the need for simplification.

Step 2: Factor the Denominator

Factor the quadratic expression in the denominator: \(x^2 - 2x - 3\). Set it equal to zero: \(x^2 - 2x - 3 = (x - 3)(x + 1)\). This shows that the denominator has a factor of \(x + 1\), which will allow for cancellation.

Step 3: Simplify the Expression

With the discovered common factor \(x + 1\), the original expression \(\frac{x+1}{(x-3)(x+1)}\) can be simplified by canceling the \(x + 1\) terms in the numerator and the denominator, resulting in \(\frac{1}{x-3}\) for \(x eq -1\).

Step 4: Evaluate the Simplified Expression at the Limit

Substitute \( x = -1 \) into the simplified expression \( \frac{1}{x-3} \). Doing so gives \( \frac{1}{-1 - 3} = \frac{1}{-4} = -\frac{1}{4} \).

Key Concepts

Indeterminate Forms

When dealing with limits in calculus, an indeterminate form often appears during substitution, specifically taking the shape of expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms indicate uncertainty about the behavior of a function as variables approach certain values. In the given exercise, plugging \( x = -1 \) into \( \frac{x+1}{x^{2}-2x-3} \) leads to \( \frac{0}{0} \), signaling that this expression is not readily solvable in its current form.

• Indeterminate forms require further manipulation or simplification.
• Common strategies include factoring, rationalizing, or applying techniques such as l’Hôpital's rule.
• Recognizing these forms is crucial to correctly evaluating limits and understanding function behavior near specific points.

Understanding and identifying indeterminate forms is a key skill in calculus, enabling a path forward in problems where limits are involved.

Limit Evaluation

Limit evaluation involves determining the value a function approaches as the input approaches a specific point. In mathematical terms, a limit describes the tendency of a function, providing essential insights into its characteristics at boundaries and critical points.The process often starts with substitution to see if the limit can be directly evaluated. If direct substitution returns an indeterminate form like \( \frac{0}{0} \), further action is required.

• Consider algebraic simplification, like factoring, to eliminate common problematic terms.
• Advanced calculus techniques, such as l’Hôpital’s rule, often apply when algebraic manipulation isn't feasible or fails.
• Successful limit evaluation reveals the behavior of a function as it nears a specific input value, aiding the understanding of continuity and potential asymptotes.

Through careful and strategic evaluation, limits provide a deeper comprehension of mathematical functions at challenging points.

Factorization

Factorization simplifies expressions and plays a crucial role in resolving indeterminate forms during limit evaluation. By breaking down complex expressions into their constituent parts, particularly polynomials, simplification and limit calculation become manageable.In our exercise, noticing that the denominator \( x^2 - 2x - 3 \) can be rephrased as \((x-3)(x+1)\) allows cancellation with the \( x+1 \) term in the numerator. Here's why factorization is vital:

• It identifies common factors that are otherwise hidden in the expression.
• Cancellation of these factors reduces complexity, eliminating indeterminate forms.
• Ultimately, it provides a pathway to evaluating limits efficiently.

Factorization is indispensable in algebra and calculus, offering clarity in expressions and making seemingly contained problems solvable.