Question

Use the following definitions. Assume fexists for all \(x\) near a with \(x>\) a. We say that 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 00\) there exists \(\delta>0\) such that $$ |f(x)-L|<\varepsilon \quad \text { whenever } \quad 0

Short answer

Question: Prove that the limit \(\lim_{x \rightarrow 0^{+}} \sqrt{x} = 0\) using the given definition of the limit. Answer: Using the definition of a limit, we showed that for any \(\varepsilon > 0\), if we choose \(\delta = \varepsilon^2\), then \(\sqrt{x} < \varepsilon\) holds for \(0 < x < \delta\). Thus, we can conclude that \(\lim_{x \rightarrow 0^{+}} \sqrt{x} = 0\).

Step-by-step solution

Since we are tasked with proving \(\lim_{x \rightarrow 0^{+}} \sqrt{x} = 0\), we need to show that for any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that the following inequality holds: $$ |\sqrt{x} - 0| < \varepsilon \quad \text{whenever} \quad 0 < x - 0 < \delta $$ #Step 2: Simplify the inequality#

We can simplify the inequality as follows: $$ |\sqrt{x}| < \varepsilon \quad \text{whenever} \quad 0 < x < \delta $$ Since both \(\sqrt{x}\) and \(\varepsilon\) are positive, the inequality can be further simplified to: $$ \sqrt{x} < \varepsilon \quad \text{whenever} \quad 0 < x < \delta $$ #Step 3: Find the relationship between \(\varepsilon\) and \(\delta\)#

To find the appropriate value for \(\delta\), we need to establish a relationship between \(\varepsilon\) and \(\delta\). We can do this by squaring both sides of the simplified inequality: $$ x < \varepsilon^2 \quad\text{whenever}\quad 0 < x < \delta $$ This implies that if \(\delta = \varepsilon^2\), then the inequality holds. #Step 4: Complete the proof#

We have shown that for any \(\varepsilon > 0\), if we choose \(\delta = \varepsilon^2\), then the inequality \(\sqrt{x} < \varepsilon\) holds for \(0 < x < \delta\). Thus, we can conclude that \(\lim_{x \rightarrow 0^{+}} \sqrt{x} = 0\).

Key concepts

Epsilon-Delta Definition

The epsilon-delta definition of a limit is a rigorous way to describe what it means for a function to approach a specific value as the input approaches a given point. This formal definition ensures precision and can be expressed mathematically.

In simple terms, for a function \( f(x) \), the statement that \( \lim_{x \to a} f(x) = L \) means the following:

• For every small distance \( \varepsilon > 0 \) you choose, isolating how close \( f(x) \) should be to \( L \),
• there exists a small interval \( \delta > 0 \) around \( a \) where all \( x \) such that \( 0 < |x - a| < \delta \)
• guarantees that the function values \( f(x) \) are within the distance \( \varepsilon \) of \( L \).

To visualize, imagine drawing a bubble around \( L \) with radius \( \varepsilon \), and finding a corresponding bubble around \( a \) of radius \( \delta \) such that within the second bubble, all the values of the function stay within the first bubble regardless of how close you examine the approach.

This approach is commonly used in calculus to define limits precisely and challenges the intuition by relying on these small thresholds of \( \varepsilon \) and \( \delta \).

Right-Hand Limit

A right-hand limit considers only the values of a function as it approaches a point from the right. Denoted as \( \lim_{x \to a^+} f(x) = L \), it signifies that the function approaches a limit \( L \) as \( x \) gets closer to \( a \) from values greater than \( a \).

For the exercise provided, we consider \( \lim_{x \to 0^+} \sqrt{x} = 0 \). This specifically means:

• \( f(x) = \sqrt{x} \)
• As \( x \to 0 \) from values greater than 0 (i.e., from the right),
• The function values \( \sqrt{x} \) get arbitrarily close to 0.

To prove this limit with the epsilon-delta definition, follow these steps:

• Select any \( \varepsilon > 0 \), which represents how close the square root function should reach the value of 0.
• Find a \( \delta > 0 \) such that if \( 0 < x < \delta \), then \( |\sqrt{x} - 0| < \varepsilon \).
• This involves setting \( \delta = \varepsilon^2 \), ensuring that \( \sqrt{x} < \varepsilon \) when \( x < \delta \).

This clearly demonstrates how right-hand limits focus on behavior coming from one direction, helping understand the function's behavior as you approach a boundary point from the right.

Left-Hand Limit

A left-hand limit is similar to a right-hand limit, but it approaches the point from the left. This means we look at the values of \( x \) that are less than \( a \), denoted as \( \lim_{x \to a^-} f(x) = L \).

Left-hand limits imply:

• For any \( \varepsilon > 0 \),
• There exists a \( \delta > 0 \) such that if \( 0 < a - x < \delta \), then \( |f(x) - L| < \varepsilon \).

Take, for example, any scenario where a function needs analysis from left approaching values. The idea is exactly the same as with right-hand limits:

• Choose a precision \( \varepsilon \) indicating the closeness to \( L \).
• Identify the corresponding boundary \( \delta \).
• Ensure within that setting, the function values stay within \( \varepsilon \) from \( L \).

While our original example highlights a right-hand limit, understanding left-hand limits applies similarly but in the opposite direction, allowing us to comprehensively dissect behavior from both sides of a point on a real-valued function.