Problem 4 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow-3} \frac{x+3}{x^{2}+2 x-3} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Indeterminate Form

First, substitute the limit value \( x = -3 \) into the expression \( \frac{x+3}{x^{2}+2x-3} \) to determine if it results in an indeterminate form. Substitute \(-3\) into the numerator and denominator:\[ \text{Numerator}: -3 + 3 = 0 \]\[ \text{Denominator}: (-3)^2 + 2(-3) - 3 = 9 - 6 - 3 = 0 \]The expression is of the form \( \frac{0}{0} \), which is indeterminate.

Step 2: Differentiate the Numerator and Denominator

Differentiate the numerator and the denominator separately with respect to \( x \):- For the numerator \( x+3 \), the derivative is \( 1 \).- For the denominator \( x^2 + 2x - 3 \), the derivative is calculated as: \[ \frac{d}{dx}(x^2 + 2x - 3) = 2x + 2 \]

Step 3: Apply l'Hospital's Rule

Use l'Hospital's Rule, which states that for limits of the form \( \frac{0}{0} \), the limit can be rewritten as:\[ \lim_{x \to -3} \frac{x+3}{x^2 + 2x - 3} = \lim_{x \to -3} \frac{1}{2x + 2} \]

Step 4: Evaluate the New Limit

Substitute \( x = -3 \) into the new limit expression obtained after applying l'Hospital's Rule:\[ 2(-3) + 2 = -6 + 2 = -4 \]Thus, the limit evaluates to:\[ \lim_{x \to -3} \frac{1}{2x + 2} = \frac{1}{-4} = -\frac{1}{4} \]

Key Concepts

Indeterminate Form

When dealing with limits in calculus, sometimes you encounter expressions that don't initially seem to have a definite value. These are known as indeterminate forms. One common indeterminate form is \( \frac{0}{0} \). This occurs when both the numerator and denominator of a fraction approach zero as \( x \) approaches a particular value. Identifying an indeterminate form is crucial because it signals that typical limit evaluation methods may not work, and that's when techniques like l'Hospital's Rule come in handy. Essentially, indeterminate forms tell us that our function is behaving in a complex way near the limit in question, requiring more advanced tools to solve. Recognizing these forms is the first step to solving limits using more sophisticated strategies like differentiation.

Differentiation

Differentiation is a method in calculus used to find the derivative of a function. The derivative measures how a function changes as its input changes. In the context of l'Hospital's Rule, differentiation allows us to simplify indeterminate forms like \( \frac{0}{0} \). By differentiating both the numerator and the denominator, we can often transform the difficult expression into one that is easier to evaluate.

• For algebraic expressions such as \( x + 3 \), differentiation is straightforward, and the derivative is \( 1 \).
• For quadratic expressions like \( x^2 + 2x - 3 \), you apply the power rule, resulting in a derivative of \( 2x + 2 \).

Differentiation reduces complex fractions into simpler ones that are simpler to solve and evaluate at specific points, especially when dealing with limits.

Limits

Limits are a fundamental concept in calculus, representing the value that a function approaches as the input approaches a certain point. In this exercise, we are interested in finding: \[ \lim_{x \to -3} \frac{x+3}{x^2 + 2x - 3} \] To solve this, we're determining what value the fraction will "settle" on as \( x \) gets infinitely close to -3. Without techniques like l'Hospital's Rule, approaching this limit directly gives us \( \frac{0}{0} \), which is indeterminate. By applying the rule and differentiating the numerator and denominator, we eventually evaluate the limit without encountering an undefined form. Limits are powerful—they help describe behavior that isn't immediately obvious, particularly when a function isn't well-defined at a specific point.

Calculus

Calculus is a branch of mathematics focusing on change. Within calculus, limits, derivatives, and integrals are key components. Limits help us understand function behavior at specific points, while differentiation and integration focus on changes and accumulation, respectively.
L'Hospital's Rule is a particularly useful technique in calculus for resolving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). By differentiating both the numerator and the denominator, you turn an unsolvable limit into one that is straightforward to compute. Calculus provides the tools to address problems of change and motion, defining behavior with precision and allowing for applications across sciences and engineering. It challenges us, but it also brings satisfaction when we solve seemingly impossible problems using logic and mathematical rigor.