Problem 6 - Calculus with Concepts in Calculus

Question

For which of the following functions can we define \(f(a)\) so as to make \(f\) continuous at \(a\) ? For those we can, find the value of \(f(a)\) that makes \(f\) continuous at \(a\). a. \(f(x)=\frac{x^{2}-9}{x-3} ; a=3\) b. \(f(x)=\frac{x^{2}+5 x+4}{x-1} ; a=1\) c. \(f(x)=\frac{e^{x}-1}{x} ; a=0\) d. \(f(x)=\sin \frac{1}{x} ; a=0\)

Step-by-Step Solution

Answer

Functions (a), (b), (c) can be made continuous at respective points by setting \( f(a) \) as 6, 5, and 1. Function (d) cannot be made continuous at \( a = 0 \).

Step 1: Determine Continuity Requirement

For a function to be continuous at a point \( a \), the limit of the function as \( x \) approaches \( a \) from both directions must equal \( f(a) \). This implies \( \lim_{{x \to a}} f(x) = f(a) \). We will explore this possibility for each function.

Step 2: Part a: Simplify and Find Limit

The function is \( f(x) = \frac{x^2 - 9}{x - 3} \). This can be simplified by factoring the numerator: \( x^2 - 9 = (x - 3)(x + 3) \). The function simplifies to \( f(x) = x + 3 \) for \( x eq 3 \). Calculate the limit: \( \lim_{{x \to 3}} (x+3) = 6 \). \( f(3) \) can be defined as 6 to make \( f \) continuous at \( x = 3 \).

Step 3: Part b: Simplify and Analyze the Limit

The function is \( f(x) = \frac{x^2 + 5x + 4}{x - 1} \). Factor the numerator: \( x^2 + 5x + 4 = (x + 1)(x + 4) \). So, \( f(x) = x + 4 \) for \( x eq 1 \). The limit is \( \lim_{{x \to 1}} (x + 4) = 5 \). \( f(1) \) can be set to 5 to make \( f \) continuous at \( x = 1 \).

Step 4: Part c: Evaluate the Limit Using L'Hospital's Rule

For \( f(x) = \frac{e^x - 1}{x} \), as \( x \to 0 \), both numerator and denominator approach 0, allowing us to use L'Hôpital's Rule. Differentiate the numerator and denominator: the derivative of \( e^x-1 \) is \( e^x \) and the derivative of \( x \) is 1. The limit becomes \( \lim_{{x \to 0}} \frac{e^x}{1} = e^0 = 1 \). Thus, setting \( f(0) = 1 \) makes \( f \) continuous at \( x = 0 \).

Step 5: Part d: Check for Limit Existence

For \( f(x) = \sin(\frac{1}{x}) \), evaluate the behavior as \( x \to 0 \). The limit \( \lim_{{x \to 0}} \sin(\frac{1}{x}) \) does not exist because \( \frac{1}{x} \) oscillates rapidly as \( x \to 0 \), causing \( \sin(\frac{1}{x}) \) to oscillate between \(-1\) and \(1\). Therefore, \( f \) cannot be made continuous at \( x = 0 \).

Key Concepts

Discontinuity

Discontinuity in a function occurs when there is a break, jump, or hole at a particular point, making it impossible to draw the function at that point without lifting the pen. At such points, the well-defined value of the function doesn't display smoothly or lacks a certain value that can be determined by its corresponding X value.

To make a function continuous at a point \( a \), the limit must exist and be equal to the function's value at that point. Here are some common types of discontinuities you might encounter:

• **Removable Discontinuity**: This happens when a function has a hole but can be 'repaired' by redefining it at that point, like in parts a and b of our exercise.
• **Jump Discontinuity**: This is when the limit from the left and the right of the point causes different values, making continuity impossible.
• **Infinite Discontinuity**: The function approaches infinity, as in asymptotes at certain points.

Understanding the type of discontinuity is crucial in deciding whether or not a function can be made continuous at a point.

Limits

The concept of limits is foundational in calculus and deals with the behavior of functions as their input approaches a particular point. A limit tells us the value that a function approaches as the input approaches some point \( x = a \). Calculating whether the function value approaches this same value from both directions (left and right) ensures understanding of continuity at that point.

In our exercise:

• For part a, the limit as \( x \to 3 \) of \( \frac{x^2 - 9}{x - 3} \) after simplifying is \( 6 \).
• In part b, as \( x \to 1 \), \( \frac{x^2 + 5x + 4}{x - 1} \)'s limit simplifies to \( 5 \).
• Part c applies L'Hôpital's Rule, finding the limit of \( \frac{e^x - 1}{x} \) as it reaches 1.
• For part d, \( \sin(\frac{1}{x}) \), the limit doesn't exist due to oscillation.

Limits help us analyze the behavior of functions at boundary points and determine if they can continue smoothly by connecting gaps for continuity.

Factorizing

Factorizing is a mathematical process used to simplify expressions, particularly when dealing with polynomials. It involves rewriting an expression as a product of its factors, thus simplifying and often removing discontinuities. This method is pivotal in reducing and simplifying fractions to evaluate limits more easily.

In our original exercise:

• In Part a, \( f(x) = \frac{x^2 - 9}{x - 3} \) was simplified after factoring the numerator: \((x - 3)(x + 3)\). Removing the \( x-3 \) factor in the denominator resolved the discontinuity.
• Part b required factoring \( x^2 + 5x + 4 \) to \((x + 1)(x + 4)\), which helped simplify the function around the point \( a = 1 \).

Through factorization, we transform complex rational functions into simpler forms so we can analyze their behavior around certain points more easily. This step ensures we remove removable discontinuities, ultimately aiding in making a discontinuous function continuous.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful mathematical tool used to calculate limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule requires taking the derivative of both numerator and denominator and recalculating the limit after differentiation. It is especially useful in cases where direct substitution doesn’t reveal a limit easily.

In this exercise:

• Part c showcases the perfect application of L'Hôpital's Rule, as \( f(x) = \frac{e^x - 1}{x} \) becomes indeterminate as \( x \to 0 \).
• By taking the derivative of the numerator (\( e^x \)) and the denominator (1), the limit simplifies to \( \lim_{{x \to 0}} e^x = 1 \).

L'Hôpital's Rule elegantly transforms complex limits into straightforward evaluations by focusing on the behavior of the derivatives at a particular point. It plays an essential role in determining the continuous nature of functions that initially appear to be discontinuous.