Problem 43 - Calculus Early Transcendentals

Question

Use the following definitions. Assume fexists for all \(x\) near a with \(x>\) a. We say the limit of \(f(x)\) as \(x\) approaches a from the right of a is \(L\) and write \(\lim _{x \rightarrow a^{+}} f(x)=L,\) if for any \(\varepsilon>0\) there exists \(\delta>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0< x-a<\delta$$ Assume fexists for all \(x\) near a with \(x < \) a. We say the limit of \(f(x)\) as \(x\) approaches a from the left of a is \(L\) and write \(\lim _{x \rightarrow a^{-}} f(x)=L,\) if for any \(\varepsilon > 0 \) there exists \(\delta > 0\) such that $$|f(x)-L| < \varepsilon \quad \text { whenever } \quad 0< a-x <\delta$$ Prove that \(\lim _{x \rightarrow 0^{+}} \sqrt{x}=0\).

Step-by-Step Solution

Answer

Question: Prove that the limit of the function \(f(x) = \sqrt{x}\) as \(x\) approaches \(0\) from the right side is equal to 0. Answer: By choosing \(\delta = \varepsilon^2\), we have shown that for any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that \(|\sqrt{x} - 0| < \varepsilon\) whenever \(0 < x < \delta\). Therefore, the limit of the function \(f(x) = \sqrt{x}\) as \(x\) approaches \(0\) from the right side is equal to 0.

Step 1: Understand the problem

We need to show that for any \(\varepsilon > 0\), there exists \(\delta > 0\) such that \(|\sqrt{x} - 0 | < \varepsilon\) whenever \(0< x < \delta\). 2.

Step 2: Choose a suitable \(\delta\)

Since we need to find a suitable \(\delta\), let's try choosing \(\delta\) in terms of \(\varepsilon\), i.e., \(\delta = \varepsilon^2\). 3.

Step 3: Show that this choice of \(\delta\) works

For \(0 < x < \delta\), we have \(0 < x < \varepsilon^2\). Then, we can apply the square root to both sides of the inequality to get \(0 < \sqrt{x} < \varepsilon\). 4.

Step 4: Calculate \(|\sqrt{x} - 0|\)

Since \(\sqrt{x}\) is positive for \(x > 0\), we have \(|\sqrt{x} - 0| = \sqrt{x}\). 5.

Step 5: Check if the given inequality is satisfied

We have \(0 < \sqrt{x} < \varepsilon\), which implies \(|\sqrt{x} - 0| = \sqrt{x} < \varepsilon\). Since we have found a suitable \(\delta = \varepsilon^2\) that satisfies the inequality \(|\sqrt{x} - 0| < \varepsilon\) for any \(\varepsilon > 0\), we can conclude that \(\lim_{x \rightarrow 0^+} \sqrt{x} = 0\).

Key Concepts

Right-hand limit

When exploring limits in calculus, it is crucial to understand the concept of the right-hand limit. A right-hand limit focuses on the behavior of a function as the variable approaches a specific point from the positive side. What this means is, you observe the values of the function as the input values increase towards that specific point, but are always slightly greater than the point itself.

A right-hand limit is denoted as \(\lim_{x \rightarrow a^+} f(x)\), where \( a \) is the point the function approaches from the right. If the function approaches a fixed number \( L \) as \( x \) gets closer to \( a \) from this direction, you can say the right-hand limit of \( f(x) \) at \( x = a \) is \( L \).

• It gives insight into the behavior of functions near discontinuities.
• Important in piecewise functions where behaviors may differ when approaching from different sides.

Consideration of one-sided limits like the right-hand limit provides a more comprehensive analysis of a function's behavior at specific points.

Delta-epsilon definition

The delta-epsilon (\( \delta-\varepsilon \)) definition is a formal method to prove limits rigorously in calculus. It provides a framework to validate that a function approaches a specific limit as the variable approaches a given point. This method is precise and avoids ambiguity, laying the foundation for precise mathematical proofs.

Here's how the delta-epsilon definition works:

• For a limit \( \lim_{x \rightarrow a} f(x) = L \), the delta-epsilon method requires that for every \( \varepsilon > 0\), no matter how small, there exists a \( \delta > 0\) such that whenever \( 0 < |x - a| < \delta\), it follows that \( |f(x) - L| < \varepsilon\).
• In this way, \( \varepsilon \) controls how close \( f(x) \) is to the limit \( L \), and \( \delta \) controls how close \( x \) needs to be to \( a \).

The delta-epsilon definition is particularly revealing when proving the limits of functions like \( \sqrt{x} \) as \( x \rightarrow 0^+ \), ensuring that every possible value within the specified interval is sufficiently close to the anticipated limit, \( L \).

Square root function

The square root function, represented as \( f(x) = \sqrt{x} \), is a fundamental element in calculus and various fields of mathematics. This function is characterized by its unique nature of only accepting non-negative inputs and producing non-negative outputs. A key aspect of \( \sqrt{x} \) is its behavior near zero, which can often be a point of inquiry in limit proofs.

Some properties of the square root function include:

• It is defined for all \( x \geq 0 \).
• It is an increasing function, meaning as \( x \) increases, \( \sqrt{x} \) also increases.
• The function approaches 0 as \( x \) approaches 0 from the right, thereby making \( \lim_{x \rightarrow 0^+} \sqrt{x} = 0 \).

Understanding the behavior of the square root function, especially how it tends towards zero, is valuable in limits and helps consolidate concepts learned through the delta-epsilon definition. This can be particularly valuable when exploring continuous and smooth phenomena, where such a function naturally arises.