Problem 17
Question
Plot the functions \(u(x), l(x)\), and \(f(x) .\) Then use these graphs along with the Squeeze Theorem to determine \(\lim _{x \rightarrow 0} f(x)\). $$ u(x)=|x|, l(x)=-|x|, f(x)=\left(1-\cos ^{2} x\right) / x $$
Step-by-Step Solution
Verified Answer
The limit \( \lim_{x \rightarrow 0} f(x) = 0 \).
1Step 1: Define the functions
We are given three functions: the upper bound function \( u(x) = |x| \), the lower bound function \( l(x) = -|x| \), and the target function \( f(x) = \frac{1 - \cos^2(x)}{x} \). We will use these for plotting and analyzing our limit.
2Step 2: Simplify the target function
Our target function is \( f(x) = \frac{1 - \cos^2(x)}{x} \). We know from trigonometric identities that \( 1 - \cos^2(x) = \sin^2(x) \), thus \( f(x) = \frac{\sin^2(x)}{x} \).
3Step 3: Plot the functions
Draw the graphs for the three functions. Function \( u(x) = |x| \) forms a V-shape with the vertex at the origin, \( l(x) = -|x| \) forms an inverted V-shape, and \( f(x) = \frac{\sin^2(x)}{x} \) oscillates in between \( u(x) \) and \( l(x) \) as \( x \) approaches 0.
4Step 4: Apply the Squeeze Theorem
To apply the Squeeze Theorem for \( x \rightarrow 0 \), we need to show that \( l(x) \leq f(x) \leq u(x) \). Knowing that \( -|x| \leq \sin^2(x) \leq |x| \), we can say \( -|x|/x \leq \sin^2(x)/x \leq |x|/x \). As \( x \to 0 \), both \( -|x|/x \) and \( |x|/x \) approach 0. Thus, by the Squeeze Theorem, \( \lim_{x \rightarrow 0} f(x) = 0 \).
Key Concepts
Limit of a functionTrigonometric identitiesFunction plotting
Limit of a function
The limit of a function, a fundamental concept in calculus, describes the behavior of a function as its input approaches a certain value. To understand how functions behave near certain points, particularly where they may not be directly defined, we use limits.
In the context of this exercise, we want to find the limit of the function \( f(x) = \frac{\sin^2{x}}{x} \) as \( x \) approaches 0. Direct substitution of 0 into \( f(x) \) would result in an indeterminate form, meaning we can't just plug in 0 and expect a meaningful answer.
The Squeeze Theorem helps us evaluate these tricky limits by bounding \( f(x) \) between two simpler functions whose limits are easily determined. This way, even if \( f(x) \) oscillates or becomes undefined at the point, we can still pin down its limiting behavior by comparing it to the neighboring boundary functions. In this problem, the conclusion from the Squeeze Theorem is \( \lim_{x \rightarrow 0} f(x) = 0 \). This result makes evaluating complex limits approachable and systematic.
In the context of this exercise, we want to find the limit of the function \( f(x) = \frac{\sin^2{x}}{x} \) as \( x \) approaches 0. Direct substitution of 0 into \( f(x) \) would result in an indeterminate form, meaning we can't just plug in 0 and expect a meaningful answer.
The Squeeze Theorem helps us evaluate these tricky limits by bounding \( f(x) \) between two simpler functions whose limits are easily determined. This way, even if \( f(x) \) oscillates or becomes undefined at the point, we can still pin down its limiting behavior by comparing it to the neighboring boundary functions. In this problem, the conclusion from the Squeeze Theorem is \( \lim_{x \rightarrow 0} f(x) = 0 \). This result makes evaluating complex limits approachable and systematic.
Trigonometric identities
Trigonometric identities are equations involving trigonometric functions that hold true for every value in their domains. They are immensely useful in simplifying expressions and solving trigonometric equations.
In this exercise, we make use of the Pythagorean identity: \(1 - \cos^2(x) = \sin^2(x)\). This identity is derived from the more familiar \( \sin^2(x) + \cos^2(x) = 1 \). By substituting \( \sin^2(x) \) for \( 1 - \cos^2(x) \) in the function \( f(x) = \frac{1 - \cos^2(x)}{x}\), the expression simplifies to \( f(x) = \frac{\sin^2(x)}{x} \).
Utilizing trigonometric identities not only simplifies our expressions but also helps us better understand and manipulate the functions involved. In the context of limits and calculus, knowing these identities allows us to transform and handle functions more effectively.
In this exercise, we make use of the Pythagorean identity: \(1 - \cos^2(x) = \sin^2(x)\). This identity is derived from the more familiar \( \sin^2(x) + \cos^2(x) = 1 \). By substituting \( \sin^2(x) \) for \( 1 - \cos^2(x) \) in the function \( f(x) = \frac{1 - \cos^2(x)}{x}\), the expression simplifies to \( f(x) = \frac{\sin^2(x)}{x} \).
Utilizing trigonometric identities not only simplifies our expressions but also helps us better understand and manipulate the functions involved. In the context of limits and calculus, knowing these identities allows us to transform and handle functions more effectively.
Function plotting
Function plotting involves graphically representing mathematical functions. It's a powerful tool for visualizing how functions behave across different values of their variable.
For the functions \( u(x) = |x| \), \( l(x) = -|x| \), and \( f(x) = \frac{\sin^2(x)}{x} \), plotting helps us see where they stand in relation to each other as \( x \) approaches 0:
For the functions \( u(x) = |x| \), \( l(x) = -|x| \), and \( f(x) = \frac{\sin^2(x)}{x} \), plotting helps us see where they stand in relation to each other as \( x \) approaches 0:
- **\( u(x) = |x| \)** - This function forms a V-shaped graph opening upwards, highlighting how the absolute value function takes only non-negative outputs.
- **\( l(x) = -|x| \)** - Similarly forms a downward V-shaped graph with negative outputs.
- **\( f(x) = \frac{\sin^2(x)}{x} \)** - Oscillates between \( l(x) \) and \( u(x) \). The fluctuations decrease as \( x \) nears zero.
Other exercises in this chapter
Problem 17
Give an \(\varepsilon-\delta\) proof of each limit fact. $$ \lim _{x \rightarrow 4} \frac{\sqrt{2 x-1}}{\sqrt{x-3}}=\sqrt{7} $$
View solution Problem 17
, find the indicated limit. In most cases, it will be wise to do some algebra first. $$ \lim _{h \rightarrow 0} \frac{(2+h)^{2}-4}{h} $$
View solution Problem 17
Find the limits. $$ \lim _{n \rightarrow \infty} \frac{n^{2}}{n+1} $$
View solution Problem 18
In Problems \(18-23\), the given function is not defined at a certain point. How should it be defined in order to make it contimuous at that point? (See Example
View solution