Question

Prove: If \(h(x) \geq 0\) for \(a

Short answer

In conclusion, we have proven that if the left-hand limit of a function \(h(x)\) exists and \(h(x) \geq 0\) for all \(x\) in the interval \((a, x_{0})\), then the left-hand limit must be greater than or equal to \(0\). Using this result, we showed that if two functions \(f_{1}(x)\) and \(f_{2}(x)\) satisfy \(f_{2}(x) \geq f_{1}(x)\) in the same interval and both left-hand limits exist, then the left-hand limit of \(f_{2}(x)\) must be greater than or equal to that of \(f_{1}(x)\).

Step-by-step solution

Analyze the function h(x) and its left-hand limit

We are given that \(h(x) \geq 0\) for \(a < x < x_{0}\), and \(\lim_{x \rightarrow x_{0}-} h(x)\) exists. To prove \(\lim_{x \rightarrow x_{0}-} h(x) \geq 0\), let's use the definition of the left-hand limit: $$ \lim_{x \rightarrow x_{0}-} h(x) = L $$ if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x\) satisfying $$ 0 < x_{0} - x < \delta \quad \text{and} \quad a < x < x_{0}, $$ we have $$ |h(x) - L| < \epsilon. $$

Use the inequality to prove the desired result

Since \(h(x) \geq 0\), we can say that \(0 \leq h(x) \leq L + \| L \|\) for \(a < x < x_{0}\). Now, assuming that \(L < 0\), we have: $$ 0 \leq h(x) - L < \epsilon $$ for \(0 < x_{0} - x < \delta \quad \text{and} \quad a < x < x_{0}\). However, if we choose \(\epsilon = - \frac{1}{2} L\), we get: $$ 0 \leq h(x) - L < - \frac{1}{2} L $$ This inequality implies $$ h(x) > \frac{1}{2} L $$ for \(0 < x_{0} - x < \delta \quad \text{and} \quad a < x < x_{0}\), which contradicts the assumption that \(L < 0\). Hence, we have proven that \(\lim_{x \rightarrow x_{0}-} h(x) \geq 0\).

Apply the result to the given functions f1(x) and f2(x)

We are given that \(f_{2}(x) \geq f_{1}(x)\) for \(a < x < x_{0}\). If both left-hand limits, $$ \lim_{x \rightarrow x_{0}-} f_{2}(x) = L_{2} \quad \text{and} \quad \lim_{x \rightarrow x_{0}-} f_{1}(x) = L_{1}, $$ exist, let's consider the function \(g(x) = f_{2}(x) - f_{1}(x)\). We know that \(g(x) \geq 0\) for \(a < x < x_{0}\). Applying the result we proved in Step 2 to function \(g(x)\), we have: $$ \lim_{x \rightarrow x_{0}-} g(x) \geq 0 $$ But, \(\lim_{x \rightarrow x_{0}-} g(x) = \lim_{x \rightarrow x_{0}-} (f_{2}(x) - f_{1}(x)) = L_{2} - L_{1}\). Thus, we have $$ L_{2} - L_{1} \geq 0 $$ which implies $$ L_{2} \geq L_{1}. $$ This means that $$ \lim_{x \rightarrow x_{0}-} f_{2}(x) \geq \lim_{x \rightarrow x_{0}-} f_{1}(x), $$ completing the proof.

Key concepts

Left-Hand Limits

In real analysis, limits help describe how a function behaves as the inputs get closer to a certain point. Specifically, the **left-hand limit** of a function at a point is concerned with how the function values behave as the inputs approach the point from the left-hand side, or as values get infinitely close from values smaller than the target point.

To understand the left-hand limit, consider a function \(h(x)\) as \(x\) approaches \(x_0\) from the left, written as \(x \to x_0^-\). The left-hand limit \(\lim_{x \to x_0^-} h(x)\) exists if, for any tiny positive number \(\epsilon\), we can find a positive \(\delta\) to ensure that every input \(x\) within \(x_0 - x < \delta\) still keeps the function \(h(x)\) within \(\epsilon\) of a particular value \(L\).

This concept is foundational in understanding continuity and limits at a point. It helps in evaluating how functions transition across different intervals and supports analyzing discontinuities effectively.

Inequalities

Inequalities are a fundamental concept in mathematics, representing relationships where one value is larger or smaller than another. In the context of limits, inequalities like \(h(x) \geq 0\) play an essential role. This relationship indicates that for all \(x\) in the specified range, \(h(x)\) takes values equal to or greater than zero.

Inequalities are employed functionally to compare limits. For example, when given that one function is bounded below by another function across a range, or when \(f_2(x) \geq f_1(x)\) for \(a < x < x_0\), inequalities help formulate and prove that the left-hand limit of \(f_2(x)\) will be greater than or equal to that of \(f_1(x)\), thus:

\[\lim_{x \to x_0^-} f_2(x) \geq \lim_{x \to x_0^-} f_1(x).\]

This use of inequalities is crucial, as it helps formalize intuition about how functions are expected to behave in relation to each other as they converge to particular values.

Function Limits

The notion of **function limits** is a cornerstone of calculus and real analysis. Limits describe the value a function approaches as its inputs move closer to a certain point. The left-hand limit is a special case focusing on inputs coming from smaller values.

Function limits are defined with the formal notation such as \(\lim_{x \to x_0^-} h(x) = L\), signifying that as \(x\) approaches \(x_0\) from the left, \(h(x)\) converges to \(L\). For such a limit to exist, we need every \[0 < x_0 - x < \delta,\]satisfies the condition:\[|h(x) - L| < \epsilon,\]for all \(\epsilon > 0\).

Function limits are applicable in diverse fields such as physics, engineering, and economics, where they model phenomena as variables approach extreme values or specific thresholds. They assist in determining continuity, the existence of derivatives, and analyzing asymptotic behavior, thus playing a crucial role in advanced mathematical understanding and application.