Problem 5 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Determine Continuity Condition

A function is continuous at a point if the limit of the function as it approaches the point from both sides is equal to the function's value at that point. For \( f(x) \) to be continuous at \( x=2 \), we need \( \lim_{{x \to 2}} f(x) = f(2) \). In this case, \( f(2) = 3 \). So, we need to show that \( \lim_{{x \to 2}} f(x) = 3 \).

Step 2: Simplify Expression for Limits

For \( x eq 2 \), \( f(x) = \frac{x^2 - x - 2}{x-2} \). We factor the numerator: \( x^2 - x - 2 = (x-2)(x+1) \). Thus, the expression simplifies to \( x+1 \) when \( x eq 2 \) and is not undefined. So, for \( x eq 2 \), \( f(x) = x+1 \).

Step 3: Calculate the Limit as x Approaches 2

Now, calculate \( \lim_{{x \to 2}} (x + 1) \). As \( x \) approaches 2, the expression \( x + 1 \) approaches 3. Therefore, \( \lim_{{x \to 2}} f(x) = \lim_{{x \to 2}} (x + 1) = 3 \).

Step 4: Confirm the Limit Equals Function Value

Since we have found that \( \lim_{{x \to 2}} f(x) = 3 \) and knowing \( f(2) = 3 \), we have \( \lim_{{x \to 2}} f(x) = f(2) \). Therefore, \( f(x) \) is continuous at \( x=2 \).

Key Concepts

Limit

One of the core concepts of continuity is the limit. To check if a function is continuous at a point, we analyze how the function behaves as it approaches that point from both sides.
Specifically, we want to ensure that the limit of the function as the input nears a certain value is the same as the function's actual value at that point.

For example, in the function given:

• As we are interested in continuity at \(x = 2\), we need to evaluate \( \lim_{{x \to 2}} f(x) \).
• Observing the condition that the value of \(f(x)\) changes at \(x=2\), the function's value is defined directly as 3.

This means our task becomes verifying if approaching 2 from either side also yields a limit of 3.
When it does, we can say that \(f(x)\) is continuous at that point.

Factorization

Factorization is a crucial algebraic tool, particularly useful when working with polynomial expressions.
It simplifies expressions, making it much easier to evaluate limits and perform continuity checks.

In this exercise, factorization of the numerator is important when simplifying:

• The original expression is \(\frac{x^2 - x - 2}{x-2}\).
• We use factorization to rewrite the numerator, \(x^2 - x - 2 = (x-2)(x+1)\).
• This allows the \((x-2)\) terms to cancel out, leaving us with \(x+1\).

By removing the indeterminate form, we now have a straightforward expression, \(f(x) = x + 1\), to evaluate as \(x\) approaches 2.

Function Value

Understanding the actual value of a function at a specific point is essential for evaluating continuity.
It acts as a check to ensure that the limits applied in both directions meet the function value specified at that point.

In our step-by-step solution:

• The value of \(f(x)\) at \(x = 2\) is given directly as 3.
• This set value is crucial as it is compared with the limit evaluated \(\lim_{{x \to 2}} f(x)\).

For \(f(x)\) to be continuous at \(x=2\), it is necessary for the limit found using the simplified expression (\(x+1\)) to match this direct function value.
When both reach the value 3 as \(x\) approaches 2, continuity is confirmed at this point.