Problem 20
Question
Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{\csc x-\cot x}{x \csc x} $$
Step-by-Step Solution
Verified Answer
The limit is 1.
1Step 1: Rewrite Trigonometric Functions to Sine and Cosine
We start by rewriting the cosecant and cotangent functions in terms of sine and cosine using the identities: - \( \csc x = \frac{1}{\sin x} \) - \( \cot x = \frac{\cos x}{\sin x} \)The limit expression becomes:\[\lim _{x \rightarrow 0} \frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x \cdot \frac{1}{\sin x}}\]Simplify the numerator:\[\frac{1 - \cos x}{\sin x}\] The entire expression simplifies to:\[\lim _{x \rightarrow 0} \frac{1 - \cos x}{x \, \sin x}\]
2Step 2: Apply Trigonometric Identity
We use the trigonometric identity for \( 1 - \cos x \): \[ 1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right) \]Substituting in the limit expression, we get:\[\lim _{x \rightarrow 0} \frac{2 \sin^2 \left(\frac{x}{2}\right)}{x \sin x}\]
3Step 3: Use Substitution and Simplify
Set \( t = \frac{x}{2} \), therefore \( x = 2t \) and as \( x \rightarrow 0, t \rightarrow 0 \).Rewriting the limit using \( t \):\[\lim _{t \rightarrow 0} \frac{2 \sin^2 t}{2t \sin(2t)}\] Simplify: \[\lim _{t \rightarrow 0} \frac{\sin^2 t}{t \cdot 2 \sin t \cos t}\] \[\lim _{t \rightarrow 0} \frac{\sin t}{t} \cdot \frac{\sin t}{2 \sin t \cos t}\]The first factor approaches 1 as \( t \rightarrow 0 \).
4Step 4: Evaluate the Simplified Limit
Focus on the remaining expression:\[\lim _{t \rightarrow 0} \frac{1}{\cos t}\]Using \( \lim_{t \to 0} \cos t = 1 \), the entire expression becomes:\[\frac{1}{1} = 1\]
5Step 5: Final Step: Conclusion of the Limit
Combining all simplified parts, we find:\[\lim _{x \rightarrow 0} \frac{\csc x - \cot x}{x \csc x} = 1\]
Key Concepts
Cosecant FunctionCotangent FunctionLimit Evaluation
Cosecant Function
The cosecant function is an essential trigonometric function often encountered in limit problems and analysis. The basic understanding revolves around its relationship with the sine function.
In mathematical terms, the cosecant function, denoted as \( \csc x \), is defined as the reciprocal of the sine function:
This simplification makes it easier to manipulate the expression and leverage identities for solving complex limit problems.
In mathematical terms, the cosecant function, denoted as \( \csc x \), is defined as the reciprocal of the sine function:
- \( \csc x = \frac{1}{\sin x} \)
- It is undefined where sine is zero, as dividing by zero is not possible. This occurs at integer multiples of \( \pi \), leading to vertical asymptotes in its graph.
- The graph of the cosecant function has a periodic nature with a period of \( 2\pi \).
- Since it is the reciprocal of the sine, whenever sine reaches its peak (1 or -1), cosecant reaches its minimum and maximum, respectively.
This simplification makes it easier to manipulate the expression and leverage identities for solving complex limit problems.
Cotangent Function
The cotangent function, similar to the cosecant function, is another handy trigonometric function, especially when dealing with limits. This function is intricately connected to both sine and cosine. In mathematical terms, the cotangent function is defined as:
This transformation is often the first step in evaluating the limit as it opens pathways to further simplification and problem-solving.
- \( \cot x = \frac{\cos x}{\sin x} \)
- It is defined wherever sine is not zero, which excludes the integer multiples of \( \pi \).
- Its periodicity is \( \pi \), indicating how the function repeats its pattern.
- Its graph consists of alternating peaks and troughs, opposite to the tangent function since it is the reciprocal.
- Cotangent reaches zero wherever cosine is zero, which aligns with multiples of \( \pi \) plus \( \frac{\pi}{2} \).
This transformation is often the first step in evaluating the limit as it opens pathways to further simplification and problem-solving.
Limit Evaluation
Limit evaluation is a powerful mathematical tool used to find the value that a function approaches as the input approaches a specific value. This concept is crucial in calculus and analysis, especially when dealing with indeterminate forms such as "0/0."
In the context of the given exercise, evaluating the limit as \( x \) approaches zero involves several strategic steps:
In this exercise, meticulous transformation and simplification lead to the limit's value, demonstrating the elegance and utility of limit evaluation in calculus problems.
In the context of the given exercise, evaluating the limit as \( x \) approaches zero involves several strategic steps:
- First, rewrite the problem using trigonometric identities. This helps simplify the expressions for easy manipulation.
- Recognize common limits, such as \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), which can be used to simplify other parts of the expression.
- Use substitution to facilitate easier evaluation of the limits. For example, substituting \( t = \frac{x}{2} \) changes the expressions to simpler forms.
- Apply trigonometric identities like \( 1 - \cos x = 2\sin^2\left(\frac{x}{2}\right) \) to further simplify and make use of known limit results.
- Finally, evaluate the simplified expression directly, often reducing to simpler components where limits are easily known.
In this exercise, meticulous transformation and simplification lead to the limit's value, demonstrating the elegance and utility of limit evaluation in calculus problems.
Other exercises in this chapter
Problem 19
In Problems \(15-24\), find the values of \(x \in \mathbf{R}\) for which the given functions are continuous. $$ f(x)=e^{-|x|} $$
View solution Problem 19
Evaluate the limits in problems. $$ \lim _{x \rightarrow \infty} \frac{3 e^{2 x}}{2 e^{2 x}-e^{x}} $$
View solution Problem 20
Use the formal definition of limits to prove each statement. $$ \lim _{x \rightarrow-\infty} \frac{x}{x+1}=1 $$
View solution Problem 20
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 0^{-}}\left(1+e^{x}\right) $$
View solution