Problem 71
Question
Find the limit, if it exists. \(\lim _{\theta \rightarrow 0} \frac{\cos \theta-1}{\theta^{2}}\)
Step-by-Step Solution
Verified Answer
The limit of the function \(\frac{\cos\theta-1}{\theta^2}\) as \(\theta\) approaches 0 is \(-\frac{1}{2}\).
1Step 1: Recognize Indeterminate Form
As \(\theta \rightarrow 0\), we have:
\(\lim _{\theta \rightarrow 0} \frac{\cos \theta - 1}{\theta^2} = \frac{\cos(0) - 1}{(0)^2} = \frac{0}{0}\)
This is an indeterminate form, so we'll use L'Hôpital's Rule to find the limit.
2Step 2: Apply L'Hôpital's Rule once
Using L'Hôpital's Rule, we differentiate the numerator and denominator with respect to \(\theta\) and compute the limit:
\(\lim _{\theta \rightarrow 0} \frac{d(\cos \theta - 1)}{d(\theta^2)} = \lim _{\theta \rightarrow 0} \frac{-\sin\theta}{2\theta}\)
However, we still have the indeterminate form 0/0, so we have to apply L'Hôpital's Rule one more time.
3Step 3: Apply L'Hôpital's Rule again
Now we differentiate both the numerator and denominator again with respect to \(\theta\) and compute the limit:
\(\lim _{\theta \rightarrow 0} \frac{d(-\sin\theta)}{d(2\theta)} = \lim _{\theta \rightarrow 0} \frac{-\cos\theta}{2}\)
4Step 4: Evaluate the Limit
Now, the limit can be easily evaluated as \(\theta\) approaches 0:
\(\lim _{\theta \rightarrow 0} \frac{-\cos \theta}{2} = \frac{-\cos(0)}{2} = \frac{-1}{2}\)
So, the limit of the given function is \(-\frac{1}{2}\).
Key Concepts
Understanding LimitsIdentifying Indeterminate FormsThe Role of Differentiation in CalculusTrigonometric Limits and Their Peculiarities
Understanding Limits
When dealing with calculus, one crucial concept to grasp is that of limits. They help examine what happens to a function as the input approaches a certain value. In this exercise, we're looking at how the function \( \frac{\cos \theta-1}{\theta^{2}} \) behaves as \( \theta \rightarrow 0 \).
Limits allow you to predict the behavior of a function without actually plugging the value directly into the equation, especially when it might result in undefined forms like \( \frac{0}{0} \). Understanding this helps in deriving insights about function continuity and whether they have finite values when approaching specific points.
To work effectively with limits, it is essential to recognize special cases or techniques that might be needed to solve them, such as L'Hôpital's Rule, often used when faced with an indeterminate form.
Limits allow you to predict the behavior of a function without actually plugging the value directly into the equation, especially when it might result in undefined forms like \( \frac{0}{0} \). Understanding this helps in deriving insights about function continuity and whether they have finite values when approaching specific points.
To work effectively with limits, it is essential to recognize special cases or techniques that might be needed to solve them, such as L'Hôpital's Rule, often used when faced with an indeterminate form.
Identifying Indeterminate Forms
Indeterminate forms are tell-tale signs in calculus computations that further evaluation is required. The most common types of indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \(0 \cdot \infty\), and others. These arise naturally when performing limit calculations. In our exercise, upon direct substitution of \( \theta = 0 \), we end up with \( \frac{0}{0} \). This calls for a special approach, such as L'Hôpital's Rule.
Recognizing these forms helps us know that the typical substitution method won't work. It requires additional manipulation of the function, using advanced calculus techniques. It's like a red flag telling us that the journey to finding the limit requires more plotting.
By identifying the indeterminate form in equations like \( \frac{\cos \theta-1}{\theta^{2}} \), it prompts the use of derivatives, offering deeper insights and correct results.
Recognizing these forms helps us know that the typical substitution method won't work. It requires additional manipulation of the function, using advanced calculus techniques. It's like a red flag telling us that the journey to finding the limit requires more plotting.
By identifying the indeterminate form in equations like \( \frac{\cos \theta-1}{\theta^{2}} \), it prompts the use of derivatives, offering deeper insights and correct results.
The Role of Differentiation in Calculus
Differentiation is a fundamental tool in calculus, describing how a function's output value changes as its input changes. It functions as the backbone of L'Hôpital's Rule, which utilizes derivatives to resolve indeterminate forms. In our example, differentiating both the numerator and denominator was necessary to find the function's limit as \( \theta \rightarrow 0 \).
Differentiation involves finding the derivative, mathematically defined as the slope of the tangent line to the function at any point. In practical applications, it tells us about the rate at which one quantity changes with respect to another. It helps in converting the original indeterminate form into a solvable expression, by reducing functions like \( \cos \theta - 1 \) and \( \theta^2 \) into terms that can be easily evaluated at the boundary points of interest.
Thus, differentiation isn't just another calculation—it's an essential process that unlocks more complex calculus ideas and resolutions.
Differentiation involves finding the derivative, mathematically defined as the slope of the tangent line to the function at any point. In practical applications, it tells us about the rate at which one quantity changes with respect to another. It helps in converting the original indeterminate form into a solvable expression, by reducing functions like \( \cos \theta - 1 \) and \( \theta^2 \) into terms that can be easily evaluated at the boundary points of interest.
Thus, differentiation isn't just another calculation—it's an essential process that unlocks more complex calculus ideas and resolutions.
Trigonometric Limits and Their Peculiarities
Trigonometric functions often present unique challenges when solving limits, due to their oscillatory nature. Consider the function \( \frac{\cos \theta-1}{\theta^{2}} \). Trigonometric limits, especially when approaching 0, often involve expressions that simplify using special techniques or identities.
In this scenario, the function \( \cos \theta \) becomes crucial as it interacts with small values of \( \theta \). Oftentimes, fundamental limits like \( \lim_{\theta \to 0} \frac{\sin\theta}{\theta}=1 \) can help, but in expressions more complex than \( \sin \theta \) or \( \cos \theta \), like this exercise entails, derivative-based methods like L'Hôpital's Rule step in to fill the gap.
These limits present unique head-scratchers, hence why having a toolbox full of different approaches is vital. Understanding and interpreting trigonometric behavior near specific points, particularly 0, is essential for smoothly navigating these types of calculus problems.
In this scenario, the function \( \cos \theta \) becomes crucial as it interacts with small values of \( \theta \). Oftentimes, fundamental limits like \( \lim_{\theta \to 0} \frac{\sin\theta}{\theta}=1 \) can help, but in expressions more complex than \( \sin \theta \) or \( \cos \theta \), like this exercise entails, derivative-based methods like L'Hôpital's Rule step in to fill the gap.
These limits present unique head-scratchers, hence why having a toolbox full of different approaches is vital. Understanding and interpreting trigonometric behavior near specific points, particularly 0, is essential for smoothly navigating these types of calculus problems.
Other exercises in this chapter
Problem 70
Let $$ f(x)=\left\\{\begin{array}{ll} -x^{2} & \text { if } x \leq 0 \\ x+1 & \text { if } x>0 \end{array}\right. $$ a. Show that \(f\) is not continuous on \([
View solution Problem 71
Let $$ f(x)=\left\\{\begin{array}{ll} -x+2 & \text { if }-2 \leq x
View solution Problem 72
Use the method of bisection to approximate the root of the equation \(x^{3}-x+1=0\) accurate to two decimal places. (Refer to Example 10.)
View solution Problem 73
Use the method of bisection to approximate the root of the equation \(x^{5}+2 x-7=0\) accurate to two decimal places.
View solution