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Chapter 2: Problem 59

Determine whether the following statements are true and give an explanation or counterexample. Assume \(a\) and \(L\) are finite numbers. a. If \(\lim _{x \rightarrow a} f(x)=L,\) then \(f(a)=L\).0 b. If \(\lim _{x \rightarrow a^{-}} f(x)=L,\) then \(\lim _{x \rightarrow a^{+}} f(x)=L\). c. If \(\lim _{x \rightarrow a} f(x)=L\) and \(\lim _{x \rightarrow a} g(x)=L,\) then \(f(a)=g(a)\). d. The limit \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\) does not exist if \(g(a)=0\). e. If \(\lim _{x \rightarrow 1^{+}} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow 1^{+}} f(x)}\), it follows that \(\lim _{x \rightarrow 1} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow 1} f(x)}\).

Short Answer

Expert verified
#Step 1# Analyze the given statements and recognize which statements are true or false. For false statements, provide counterexamples to demonstrate how it is false. #Step 2# Summarize the discovered information for each statement. a. Statement A is False. Counterexample: f(x) = (x^2 - 1)/(x-1); lim_(x鈫�1) f(x) = 2, but f(1) is not defined. b. Statement B is False. Counterexample: f(x) = 1/x; left-hand limit as x鈫�0 is -鈭�, right-hand limit as x鈫�0 is +鈭�. c. Statement C is False. Counterexample: f(x) = x and g(x) = (x^2 - 1)/(x-1); lim_(x鈫�1) f(x) = lim_(x鈫�1) g(x) = 1, but f(1) = 1 and g(1) is not defined. d. Statement D is False. Counterexample: f(x) = sin(x) and g(x) = x; lim_(x鈫�0) (sin(x)/x) = 1, even though g(0) = 0. e. Statement E: Cannot conclude True or False without more information about the left-hand limit and right-hand limit of f(x) at x = 1. If the right-hand limit and left-hand limit are the same, then the statement is True. Otherwise, the conclusion may be invalid.

Step by step solution

01

Statement A: The relationship between the limit and the function's value at the limit point

If \(lim_{x\rightarrow a} f(x) = L,\) this means that the function \(f(x)\) approaches the value \(L\) as \(x\) approaches \(a\). It does not necessarily imply that \(f(a) = L\). Counterexample: Consider the function \(f(x) = \frac{x^2 - 1}{x-1}.\) We can easily see that \(\lim_{x\rightarrow 1} f(x) = 2\), but \(f(1)\) is not defined.
02

Statement B: The relationship between left and right-hand limits

If \(\lim_{x\rightarrow a^{-}} f(x) = L,\) then the left-hand limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\). It does not necessarily mean that the right-hand limit, \(\lim_{x\rightarrow a^{+}} f(x) = L\). Counterexample: Consider the function \(f(x) = \frac{1}{x}\). The left-hand limit as \(x \rightarrow 0\) is \(-\infty\) whereas the right-hand limit is \(+\infty\).
03

Statement C: Equality of two functions when their limits are equal

If \(\lim_{x\rightarrow a} f(x) = L\) and \(\lim_{x\rightarrow a} g(x) = L\), it means that both functions approach the same value \(L\) as \(x\) approaches \(a\). However, it does not guarantee that \(f(a) = g(a)\). Counterexample: Consider the functions \(f(x) = x\) and \(g(x) = \frac{x^2 - 1}{x - 1}.\) We have \(\lim_{x\rightarrow 1} f(x) = 1\) and \(\lim_{x\rightarrow 1} g(x) = 1\), but \(f(1) = 1\) and \(g(1)\) is not defined.
04

Statement D: The limit of the ratio of functions when the denominator is zero

If \(g(a) = 0\), we cannot say that the limit \(\lim_{x\rightarrow a} \frac{f(x)}{g(x)}\) does not exist without analyzing the function further. Even though the denominator is zero at the point, it does not necessarily mean that the limit does not exist. It might be an indeterminate form like \(0/0\). Counterexample: Consider the function \(\frac{\sin(x)}{x}\) as \(x \rightarrow 0\). We have \(g(0) = 0\), but \(\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1\).
05

Statement E: Limit inside a root function

If \(\lim_{x\rightarrow 1^{+}} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow 1^{+}} f(x)}\), it follows that the square root function is continuous, and we can move the limit inside or outside the square root function. However, we cannot conclude that \(\lim _{x \rightarrow 1} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow 1} f(x)}\) unless we know that the right-hand limit and the left-hand limit are the same i.e., \(\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)\). Only then can we claim that the overall limit exists and is equal to that value.

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Key Concepts

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

Left-Hand Limit
In calculus, understanding the concept of a left-hand limit is crucial for analyzing the behavior of functions as inputs approach a particular point. A left-hand limit refers specifically to the value that a function approaches as the input variable approaches a certain value, a, from the left side or from values less than a. Mathematically, it is denoted as \( \lim_{x \rightarrow a^{-}} f(x) = L \), where L is the left-hand limit.

It is important to understand that the left-hand limit may be different from the right-hand limit at the same point, which implies the overall limit might not exist there. For example, with the function f(x) = 1/x, the left-hand limit as x approaches 0 is -鈭�, showcasing how functions can behave quite differently from either side of a point.
Right-Hand Limit
Complementary to the left-hand limit, the right-hand limit is essential for examining what happens to a function as the input comes from values greater than a particular point, a. It is denoted as \( \lim_{x \rightarrow a^{+}} f(x) = L \), where L represents the limit from the right.

A key point to remember is that for the overall limit at a point to exist, both the left-hand and right-hand limits must be equal. If they are not, this disparity may result in a discontinuity. For instance, if we look again at f(x) = 1/x, the right-hand limit as x approaches 0 is +鈭�, contrasting with the left-hand limit and indicating there is no overall limit at x=0.
Limit of a Function
The limit of a function is a fundamental concept in calculus, capturing the idea of approaching a certain value. When we say \( \lim_{x\rightarrow a} f(x) = L \), we imply that as x gets closer and closer to a from either side, f(x) approaches the limit L.

It's vital to recognize that the limit does not concern itself with the actual value of the function at a; it is solely about the value that the function is approaching. Moreover, the function need not be defined at the point a for the limit to exist. This serves as a reminder that understanding limits is about behavior near a point rather than at that point.
Indeterminate Forms
In the realm of limits, indeterminate forms are expressions that do not readily reveal the limit of a function. These forms pose a challenge because they do not clearly show whether a limit exists or what value it may have. Common indeterminate forms include \(0/0\), \(\infty/\infty\), \(0\cdot\infty\), \(\infty - \infty\), \(1^\infty\), and others.

To determine the limits of functions that yield indeterminate forms, one can often employ algebraic manipulation, L'H么pital's rule, or other calculus tools. For example, the function \(\frac{\sin(x)}{x}\) gives us the indeterminate form \(0/0\) as x approaches 0. By applying L'H么pital's rule or using the special trigonometric limit, we can find <\br> that the limit is actually 1, illustrating that these forms require careful analysis to find the underlying limits.

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Most popular questions from this chapter

Use analytical methods and/or a graphing utility en identify the vertical asymptotes (if any) of the following functions. $$f(x)=\frac{1}{\sqrt{x} \sec x}$$

A monk set out from a monastery in the valley at dawn. He walked all day up a winding path, stopping for lunch and taking a nap along the way. At dusk, he arrived at a temple on the mountaintop. The next day the monk made the return walk to the valley, leaving the temple at dawn, walking the same path for the entire day, and arriving at the monastery in the evening. Must there be one point along the path that the monk occupied at the same time of day on both the ascent and descent? (Hint: The question can be answered without the Intermediate Value Theorem.) (Source: Arthur Koestler, The Act of Creation.)

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