Problem 49 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that \(f(x) g(x)=1\) for all \(x\) and \(\lim _{x \rightarrow a} g(x)=0\). Prove that \(\lim _{x \rightarrow a} f(x)\) does not exist.

Step-by-Step Solution

Answer

\( \lim_{x \rightarrow a} f(x) \) does not exist due to a contradiction with the limit product rule.

Step 1: Understand the Given Conditions

We know that \( f(x) \cdot g(x) = 1 \) for all \( x \). We also know \( \lim_{x \rightarrow a} g(x) = 0 \). We are asked to prove that \( \lim_{x \rightarrow a} f(x) \) does not exist.

Step 2: Analyze \( f(x) \) Using Known Limits

Using the limit property of products, if \( \lim_{x \rightarrow a} [f(x) \cdot g(x)] = \lim_{x \rightarrow a} 1 = 1 \), then \( \lim_{x \rightarrow a} f(x) \cdot \lim_{x \rightarrow a} g(x) = 1 \). However, since \( \lim_{x \rightarrow a} g(x) = 0 \), the product of the limits should be \( 0 \), creating a contradiction.

Step 3: Contradiction Analysis

We see that \( \lim_{x \rightarrow a} f(x) \cdot 0 = 1 \), which implies \( f(x) \cdot 0 \) cannot equal \( 1 \) unless \( \lim_{x \rightarrow a} f(x) \) was undefined or approaching infinity, leading to a contradiction. Thus, \( \lim_{x \rightarrow a} f(x) \) must not exist.

Key Concepts

Product of functions

When dealing with functions in calculus, understanding how they work together is crucial, especially when they are multiplied. The product of functions involves taking two functions and creating a new function by multiplying their outputs. If you have functions \( f(x) \) and \( g(x) \), their product is denoted as \( f(x) \cdot g(x) \).Let's consider a simple scenario: suppose \( f(x) = 2x \) and \( g(x) = 3x+1 \). The product \( h(x) = f(x) \cdot g(x) \) would be \( 2x(3x+1) \), which simplifies to \( 6x^2 + 2x \). This is a new function whose graph and behavior reflect the interplay between \( f(x) \) and \( g(x) \).Why is this important? In calculus, products are everywhere. We find them in derivatives, integrals, and especially limits. Recognizing how individual functions affect each other through multiplication gives us insight into more complex problems.

Limit of a function

The concept of the limit is foundational in calculus. It captures the idea of a function's output as the input approaches a particular value. The notation \( \lim_{x \to a} f(x) \) signifies the value that \( f(x) \) approaches as \( x \) gets infinitely close to \( a \).Imagine you're watching a graph. As you zero in on a point from both sides, the value you're honing in on is the limit. For instance, consider the function \( f(x) = \frac{x^2 - 1}{x - 1} \). At \( x = 1 \), this function is undefined, but if we evaluate \( \lim_{x \to 1} f(x) \), it equals 2. This is because if you plug in values close to 1 into the equation, the outputs approach 2.This magic of the limit helps us navigate through discontinuities and define derivatives. It acts as a stepping stone towards understanding the behavior of functions near particular points, even if the function is not defined at that exact point.

Existence of limits

Determining whether a limit exists is pivotal. Not every function behaves nicely as we approach a point. A limit exists if, as we approach a certain \( x = a \), both from the left and the right, we settle on the same number.Consider the guidance given by limits. If we have \( \lim_{x \to a} f(x) = L \), it implies both the left-hand limit (as \( x \to a^- \)) and the right-hand limit (as \( x \to a^+ \)) are \( L \). This agreement is essential; any discrepancy implies no limit exists.In the context of our exercise, we faced a contradiction. Despite having \( \lim_{x \to a}[f(x) \cdot g(x)] = 1 \), the fact that \( \lim_{x \to a} g(x) = 0 \) disrupted this unity, given the multiplication properties of limits. Therefore, \( \lim_{x \to a} f(x) \) must not exist because no consistent \( L \) can satisfy both sides of the situation described.