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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

Expert verified
Question: Prove the limits for the following piecewise function: \(f(x) = \begin{cases} 2x - 4 & \text{if } x \geq 0 \\ 3x - 4 & \text{if } x < 0 \end{cases}\) (a) \(\lim_{x \rightarrow 0^+} f(x) = -4\) (b) \(\lim_{x \rightarrow 0^-} f(x) = -4\) (c) \(\lim_{x \rightarrow 0} f(x) = -4\) Answer: (a) For the right-sided limit, we analyzed the function for \(x \geq 0\) and proved that \(\lim_{x \rightarrow 0^+} f(x) = -4\). (b) For the left-sided limit, we analyzed the function for \(x < 0\) and proved that \(\lim_{x \rightarrow 0^-} f(x) = -4\). (c) Since both right-sided and left-sided limits are equal to -4, the two-sided limit exists and is equal to -4. Thus, \(\lim_{x \rightarrow 0} f(x) = -4\).

Step by step solution

01

Determine the function for x approaching 0 from the right

Since we are looking at \(x\) approaching \(0\) from the right side, we will use the part of the function defined for \(x \geq 0\): \(f(x) = 2x - 4\).
02

Select an arbitrary ε

Let \(\varepsilon > 0\) be arbitrary.
03

Set up the inequality we need to satisfy

We want to find a \(\delta > 0\) such that \(|2x - 4 - (-4)| < \varepsilon\) whenever \(0 < x - 0 < \delta\).
04

Simplify the inequality

Our inequality simplifies to \(|2x| < \varepsilon\). Now, we need to find the appropriate \(\delta\).
05

Find the corresponding δ

Dividing both sides of the inequality by \(2\), we get \(|x| < \frac{\varepsilon}{2}\). Since we need to satisfy \(0 < x < \delta\), we set \(\delta = \frac{\varepsilon}{2}\). Thus, we have proved \(\lim_{x \rightarrow 0^+} f(x) = -4\). #b. Proving the left-sided limit#
06

Determine the function for x approaching 0 from the left

Since we are looking at \(x\) approaching \(0\) from the left side, we will use the part of the function defined for \(x < 0\): \(f(x) = 3x - 4\).
07

Select an arbitrary ε

Let \(\varepsilon > 0\) be arbitrary.
08

Set up the inequality we need to satisfy

We want to find a \(\delta > 0\) such that \(|3x - 4 - (-4)| < \varepsilon\) whenever \(0 < 0 - x < \delta\).
09

Simplify the inequality

Our inequality simplifies to \(|3x| < \varepsilon\). Now, we need to find the appropriate \(\delta\).
10

Find the corresponding δ

Dividing both sides of the inequality by \(3\), we get \(|x| < \frac{\varepsilon}{3}\). Since we need to satisfy \(0 < -x < \delta\), we set \(\delta = \frac{\varepsilon}{3}\). Thus, we have proved \(\lim_{x \rightarrow 0^-} f(x) = -4\). #c. Proving the two-sided limit#
11

Check the right-sided and left-sided limits

We have proved that the right-sided limit and left-sided limit are both equal to \(-4\).
12

Use the definition of the two-sided limit

By definition, since the right-sided limit and the left-sided limit are equal, the two-sided limit exists and is equal to the common value, which is \(-4\). Thus, we have proved \(\lim_{x \rightarrow 0} f(x) = -4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right-hand limit
When dealing with limits, the right-hand limit is an important concept. This refers to the limit of a function as the variable approaches a certain value from the right side, which means considering values greater than the point of interest. In mathematical notation, the right-hand limit of a function \( f(x) \) as \( x \) approaches \( a \) from the right is written as \( \lim_{x \to a^+} f(x) = L \).

To determine this limit, we assume \( x \) can be slightly greater than \( a \) and check if \( f(x) \) tends to a number \( L \) for values of \( x \) close to \( a \). We achieve this by using the \( \varepsilon - \delta \) definition:
  • Choose any small positive number \( \varepsilon \).
  • Find a \( \delta > 0 \) such that \( |f(x) - L| < \varepsilon \) whenever \( 0 < x - a < \delta \).
In the exercise, when proving the limit \( \lim_{x \to 0^+} f(x) = -4 \), we use the function part \( 2x - 4 \) valid for \( x \geq 0 \). This means that any small positive deviation from zero confirms the function's limit approaching \(-4\) from the right.
Left-hand limit
The left-hand limit focuses on approaching a particular point from values less than the point itself. When finding the left-hand limit of a function \( f(x) \) as \( x \) approaches \( a \) from the left, it is denoted as \( \lim_{x \to a^-} f(x) = L \).

This concept is pivotal when examining the behavior of a function as \( x \) gets closer to \( a \) from the left side. Like the right-hand limit, it uses the \( \varepsilon - \delta \) approach:
  • Select an arbitrary small positive \( \varepsilon \).
  • Determine a \( \delta > 0 \) ensuring that \( |f(x) - L| < \varepsilon \) whenever \( 0 < a - x < \delta \).
In the given exercise, to show \( \lim_{x \to 0^-} f(x) = -4 \), we use \( 3x - 4 \) defined for \( x < 0 \). This helps observe the function behaving consistently with \(-4\) as \( x \) nears zero from the left, ensuring that \( f(x) \) approaches \(-4\) accurately through this approach.
Two-sided limit
The two-sided limit evaluates how a function behaves as it approaches a point from both the left and right sides simultaneously. For a two-sided limit to exist at a point \( a \), both the right-hand limit \( \lim_{x \to a^+} f(x) \) and the left-hand limit \( \lim_{x \to a^-} f(x) \) must exist and be equal. This is usually denoted simply as \( \lim_{x \to a} f(x) = L \).

Understanding this concept is crucial, allowing one to understand whether a function is continuous at a specific point or if there is a common behavior from both directions.
  • Confirm the existence and equality of both one-sided limits.
  • The two-sided limit \( \lim_{x \to a} f(x) = L \) exists if these conditions are met.
In the scenario of the exercise, confirmation of \( \lim_{x \to 0} f(x) = -4 \) is achieved because both one-sided limits \( \lim_{x \to 0^+} f(x) = -4 \) and \( \lim_{x \to 0^-} f(x) = -4 \) are equal, signifying the function's value approaches \(-4\) regardless of the direction from which \( x \) approaches zero.

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