/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Let \(f(x)=\frac{|x|}{x},\) for ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 36

Let \(f(x)=\frac{|x|}{x},\) for \(x \neq 0\) a. Sketch a graph of \(f\) on the interval [-2,2] b. Does \(\lim _{x \rightarrow 0} f(x)\) exist? Explain your reasoning after first examining \(\lim _{x \rightarrow 0^{-}} f(x)\) and \(\lim _{x \rightarrow 0^{+}} f(x)\)

Short Answer

Expert verified
Answer: The function \(f(x)\) is equal to 1 for \(x>0\), and is equal to -1 for \(x<0\). The graph has a hole at x=0, and the limit as x approaches 0 does not exist.

Step by step solution

01

Evaluate f(x) for x>0 and x

Start by analyzing the behavior of the function for two different cases: 1. When \(x>0\), the absolute value of \(x\) is just \(x\), so we have \(f(x)=\frac{x}{x}=1\). 2. When \(x<0\), the absolute value of \(x\) is \(-x\), so we have \(f(x)=\frac{-x}{x}=-1\).
02

Sketch the graph

Since we have found out that \(f(x)=1\) for \(x>0\) and \(f(x)=-1\) for \(x<0\), we can now sketch the graph of the function on the interval [-2, 2]. Note that the graph will have a hole at x=0 since the function is not defined there.
03

Compute the left-sided limit

We need to find the left-sided limit, which is \(\lim_{x\rightarrow 0^-} f(x)\). We know that for \(x<0\), \(f(x)=-1\). As \(x\) approaches 0 from the left side, it is still negative, so the left-sided limit is -1: \(\lim_{x\rightarrow 0^-} f(x)=-1\).
04

Compute the right-sided limit

Now, we need to find the right-sided limit, which is \(\lim_{x\rightarrow 0^+} f(x)\). We know that for \(x>0\), \(f(x)=1\). As \(x\) approaches 0 from the right side, it is still positive, so the right-sided limit is 1: \(\lim_{x\rightarrow 0^+} f(x)=1\).
05

Determine if the overall limit exists

We analyzed both the left-sided and right-sided limits and found out that \(\lim_{x\rightarrow 0^-} f(x)=-1\) and \(\lim_{x\rightarrow 0^+} f(x)=1\). Since these two limits are not equal, the overall limit \(\lim_{x\rightarrow 0} f(x)\) does not exist. Thus, the graph of \(f(x)\) has a hole at x=0, is equal to 1 for x>0, and is equal to -1 for x<0, and the limit as x approaches 0 does not exist.

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Key Concepts

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

Piecewise Functions
Piecewise functions are a special type of function that is defined by multiple sub-functions, each applicable to a certain part of the domain. This means that the function behaves differently over various intervals. For example, with the function \( f(x) = \frac{|x|}{x} \), it can be broken down into parts, based on the sign of the input \( x \).
  • When \( x > 0 \), the function simplifies to \( f(x) = 1 \).
  • When \( x < 0 \), it simplifies to \( f(x) = -1 \).
  • And for \( x = 0 \), the function is undefined because division by zero is not possible, which creates a hole in the graph at this point.
The behavior of a piecewise function like \( f(x) \) allows it to approach different values on either side of the undefined point. Understanding piecewise functions means focusing on how each piece behaves and transitions across its domain.
One-Sided Limits
One-sided limits help us understand how a function behaves as the input \( x \) approaches a particular value from one direction only. This concept is especially useful for piecewise functions or functions with breaks, like \( f(x) = \frac{|x|}{x} \).
  • The **left-sided limit** \( \lim_{x \rightarrow 0^-} f(x) \) considers values approaching from the left side of zero. Since for \( x < 0 \), \( f(x) = -1 \), the left-sided limit is \( -1 \).
  • The **right-sided limit** \( \lim_{x \rightarrow 0^+} f(x) \) considers values from the right side of zero. For \( x > 0 \), \( f(x) = 1 \), making the right-sided limit \( 1 \).
By understanding one-sided limits, you can determine the behavior of functions at points where standard limits might not exist. These limits emphasize that sometimes a function behaves differently when approached from different sides.
Discontinuous Functions
A function is considered discontinuous at a certain point if it has a break, jump, or hole at that point. The function \( f(x) = \frac{|x|}{x} \) is discontinuous at \( x = 0 \) due to different behaviors on either side of the zero.
  • The graph of \( f(x) \) illustrates a jump discontinuity because the left-side (\( x < 0 \), \( f(x) = -1 \)) doesn't connect smoothly to the right-side (\( x > 0 \), \( f(x) = 1 \)).
  • Since \( \lim_{x \rightarrow 0^-} f(x) = -1 \) and \( \lim_{x \rightarrow 0^+} f(x) = 1 \), the limit at \( x = 0 \) does not exist.
Such discontinuities reveal important features about how a function behaves and whether it can be graphed without lifting the pencil off the paper. Discontinuous points like these are crucial in calculus and analysis, helping to define the nature of the function's graph.

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Most popular questions from this chapter

\(A\) function \(f\) is even if \(f(-x)=f(x)\) for all \(x\) in the domain of \(f\). Suppose \(f\) is even. with \(\lim _{x \rightarrow 2^{+}} f(x)=5\) and \(\lim _{x \rightarrow 2^{-}} f(x)=8 .\) Evaluate the following limits. a. \(\lim _{x \rightarrow-2^{+}} f(x)\) b. \(\lim _{x \rightarrow-2^{-}} f(x)\)

Calculate the following limits using the factorization formula \(x^{n}-a^{n}=(x-a)\left(x^{n-1}+x^{n-2} a+x^{n-3} a^{2}+\cdots+x a^{n-2}+a^{n-1}\right)\) where \(n\) is a positive integer and a is a real number. $$\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{x-16}$$

A cylindrical tank is filled with water to a depth of 9 meters. At \(t=0,\) a drain in the bottom of the tank is opened and water flows out of the tank. The depth of water in the tank (measured from the bottom of the tank) \(t\) seconds after the drain is opened is approximated by \(d(t)=(3-0.015 t)^{2},\) for \(0 \leq t \leq 200 .\) Evaluate and interpret \(\lim _{t \rightarrow 200^{-}} d(t)\).

The limit at infinity \(\lim _{x \rightarrow \infty} f(x)=L\) means that for any \(\varepsilon>0\) there exists \(N>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad x>N.$$ Use this definition to prove the following statements. $$\lim _{x \rightarrow \infty} \frac{10}{x}=0$$

Use the following definition for the nonexistence of a limit. Assume \(f\) is defined for all values of \(x\) near a, except possibly at a. We write \(\lim _{x \rightarrow a} f(x) \neq L\) if for some \(\varepsilon>0\) there is no value of \(\delta>0\) satisfying the condition $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0<|x-a|<\delta.$$ Let $$f(x)=\left\\{\begin{array}{ll} 0 & \text { if } x \text { is rational } \\ 1 & \text { if } x \text { is irrational. } \end{array}\right.$$ Prove that \(\lim _{x \rightarrow a} f(x)\) does not exist for any value of \(a\). (Hint: Assume \(\lim _{x \rightarrow a} f(x)=L\) for some values of \(a\) and \(L\) and let \(\varepsilon=\frac{1}{2}\).)
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