Problem 5 - Calculus for Biology and Medicine

Question

Show that $$ f(x)=\left\\{\begin{array}{cl} \frac{x^{2}-x-2}{x-2} & \text { if } x \neq 2 \\ 3 & \text { if } x=2 \end{array}\right. $$ is continuous at \(x=2\).

Step-by-Step Solution

Answer

The function \( f(x) \) is continuous at \( x = 2 \) because \( f(2) = 3 \) and \( \lim_{x \to 2} f(x) = 3 \).

Step 1: Define Continuity at a Point

A function \( f(x) \) is continuous at a point \( x = a \) if the following three conditions are satisfied: 1. \( f(a) \) is defined. 2. \( \lim_{x \to a} f(x) \) exists. 3. \( \lim_{x \to a} f(x) = f(a) \).

Step 2: Check Function Value at \( x=2 \)

For the function \( f(x) \), we are given that if \( x = 2 \), then \( f(x) = 3 \). Thus, \( f(2) = 3 \) is defined.

Step 3: Calculate the Limit as \( x \to 2 \)

For \( x eq 2 \), \( f(x) = \frac{x^2 - x - 2}{x - 2} \). To find the limit as \( x \to 2 \), simplify the expression:\(\frac{x^2 - x - 2}{x - 2} = \frac{(x-2)(x+1)}{x-2}\)When \( x eq 2 \), \( x-2 \) can be cancelled out, giving \( x + 1 \). Therefore, \( \lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 1) = 2 + 1 = 3 \).

Step 4: Confirm Continuity Conditions

We found that \( \lim_{x \to 2} f(x) = 3 \) and \( f(2) = 3 \). Since these are equal, the function is continuous at \( x = 2 \).

Key Concepts

Limits of Functions

A limit in calculus is a way to describe the behavior of a function as it approaches a certain point. Understanding limits is fundamental to grasping concepts of continuity, derivatives, and integrals.
For a function \( f(x) \) to have a limit as \( x \) approaches a number \( a \), the values of \( f(x) \) must get arbitrarily close to a single number, which is the limit value, as \( x \) approaches \( a \) from both sides.
In the exercise provided, we are interested in the limit of the piece \( \frac{x^2 - x - 2}{x - 2} \) as \( x \) approaches 2. This expression is undefined at \( x = 2 \) due to division by zero.
However, limits allow us to explore how \( f(x) \) behaves near \( x = 2 \) by simplifying it to \( x + 1 \) after canceling the common factor \( x-2 \). Thus, as \( x \to 2 \), \( x+1 \to 3 \), which shows us the limit exists and equals 3.

Piecewise Functions

Piecewise functions are special functions defined by multiple sub-functions, each applying to a specific interval of the domain.
In our exercise, the function \( f(x) \) is defined as a piecewise function:

• For \( x eq 2 \), the function is given by \( \frac{x^2 - x - 2}{x - 2} \).
• For \( x = 2 \), it is defined as \( f(x) = 3 \).

Piecewise functions are beneficial when dealing with scenarios where a single formula is inadequate to describe a function over its entire domain.
To verify continuity at a point for piecewise functions, it is essential to confirm that all the pieces \( f(x) \) behave the same as \( x \) approaches the point. Here, simplifying \( \frac{x^2 - x - 2}{x - 2} \) shows that it approaches 3, aligning with \( f(2) = 3 \), fulfilling the requirements of continuity at \( x = 2 \).

Evaluating Limits

Evaluating limits involves determining the value a function approaches as the input nears a specified point. This is crucial, especially when direct substitution leads to indeterminate forms like \( \frac{0}{0} \).To evaluate a limit:

• Directly substitute the number into \( f(x) \) unless it causes an undefined expression.
• If undefined, simplify \( f(x) \), often by factoring or rationalizing.
• Once simplified, substitute the value again to find the limit.

In the exercise, substituting \( x = 2 \) initially leads to \( \frac{0}{0} \). Simplifying \( \frac{x^2 - x - 2}{x - 2} \) to \( x + 1 \) lets us substitute 2 for \( x \), resulting in a value of 3.
Each of these steps demonstrates the fundamental approach to evaluating limits, ensuring both existence and continuity at the specified point.