Continuity Of Functions: A Detailed Guide

by TextBrain Team 42 views

Hey guys, let's dive into a classic math problem! We're going to explore the concept of continuity in functions. This is super important stuff in calculus and understanding how functions behave. We'll be looking at two specific examples, breaking them down step-by-step so you can totally grasp the concepts. Our goal is to determine if the function is continuous at certain points, and we'll use the definition of continuity to do it. Let's get started!

Understanding Continuity: The Basics

Before we get to the nitty-gritty, let's refresh our memory on what continuity actually means. Basically, a function is continuous at a point if you can draw its graph without lifting your pen. That means no jumps, holes, or breaks. Formally, a function f(x) is continuous at a point 'a' if three conditions are met:

  1. The function must be defined at 'a': This means f(a) exists.
  2. The limit of the function as x approaches 'a' must exist: This means lim (x→a) f(x) exists.
  3. The limit and the function's value must be equal: This means lim (x→a) f(x) = f(a).

If any of these conditions fail, the function is discontinuous at that point. Keep in mind that continuity can also be explored over intervals – a function is continuous over an interval if it's continuous at every point within that interval. This is super important in many real-world applications, for example, modeling physical phenomena.

When we say a function is continuous, we are essentially saying that the function's output changes smoothly as its input changes. If there's a sudden jump or a hole in the graph, the function is not continuous at that point. This is why the concept of limits is so crucial because they help us analyze the behavior of a function near a point, even if the function isn't defined at that exact point.

Continuity is not only a fundamental concept in calculus but also has deep implications in various fields like physics, engineering, and computer science. It ensures that the models we use to describe real-world phenomena behave predictably. It also helps to analyze the behavior of functions at specific points.

Case a: Examining the Continuity of f(x) = (x² - 1 + √(x+1))/x

Let's get down to business with our first function, which is defined as:

f(x) = (x² - 1 + √(x+1)) / x,   if x ≠ 0
f(0) = 1/2,                     if x = 0

Our goal is to figure out if this function is continuous, especially around x = 0. First, we check if the function is defined at x = 0. We're given that f(0) = 1/2, so that part is covered. Next, we need to see if the limit exists as x approaches 0. That means we must evaluate:

lim (x→0) (x² - 1 + √(x+1)) / x

To evaluate this limit, we can try a few things. Plugging in x = 0 directly gives us an indeterminate form (0/0), so let's try a different approach. We can try using L'Hopital's Rule, which states that if the limit of f(x)/g(x) is indeterminate, then the limit is equal to the limit of f'(x)/g'(x). However, to make things easier, let's rationalize the numerator a bit. Multiply the numerator and denominator by the conjugate of the numerator (x² - 1 + √(x+1)): we will get something that is easier to deal with.

However, before going any further, we need to address the domain of the function. The term √(x+1) requires that x + 1 ≥ 0, meaning x ≥ -1. The denominator cannot be zero (x ≠ 0). Thus, the domain D_f of the function is [-1, 0) ∪ (0, ∞).

When rationalizing, it can be quite complex. An easier option would be to use the Taylor expansion of the square root term around x = 0. Using the Taylor expansion for √(1 + x) around x = 0, which is 1 + x/2 - x²/8 + ...,

we can rewrite our function when x is close to zero as:

f(x) ≈ (x² - 1 + 1 + x/2) / x
f(x) ≈ (x² + x/2) / x
f(x) ≈ x + 1/2

So, it looks like the limit as x approaches 0 is 1/2. Let's confirm this using the limit: lim (x→0) (x + 1/2) = 1/2.

Now, compare the value of the function at x = 0, which is f(0) = 1/2, and the limit we've just found, which is 1/2. The two values match! Therefore, the function is continuous at x = 0.

Case b: Further Analysis of Continuity

Unfortunately, we don't have the definition for case b. However, let's discuss how we would approach a similar problem, so you're ready for anything.

The Strategy for Analyzing Continuity

The approach is always the same:

  1. Check the Domain: What values of 'x' are allowed? Look for any restrictions (e.g., division by zero, square roots of negative numbers, logarithms of non-positive numbers).
  2. Evaluate at the Point in Question: Does f(a) exist? If not, the function is immediately discontinuous at 'a'.
  3. Find the Limit: Calculate lim (x→a) f(x). This might involve direct substitution, factoring, rationalizing, or using L'Hopital's Rule.
  4. Compare and Conclude: If f(a) and lim (x→a) f(x) both exist and are equal, the function is continuous at 'a'. If not, it is discontinuous.

Dealing with Different Types of Discontinuity

There are a few different kinds of discontinuity you might encounter:

  • Removable Discontinuity: This happens when there's a