Chapter 1: Problem 46
Use a calculator to graph the function and estimate the value of the limit, then use L'Hôpital's rule to find the limit directly. $$ \lim _{x \rightarrow \pi} \frac{1+\cos x}{\sin x} $$
Short Answer
Expert verified
The limit is 0.
Step by step solution
01
Graph the Function
First, use a graphing calculator or software to graph the function \( f(x) = \frac{1 + \cos x}{\sin x} \). Look at the behavior of the function as \( x \) approaches \( \pi \). You should notice that it appears the function approaches a certain value as \( x \to \pi \).
02
Identify the Indeterminate Form
Substitute \( x = \pi \) into the given function. We get \( \frac{1 + \cos \pi}{\sin \pi} = \frac{1 - 1}{0} = \frac{0}{0} \). Since this results in an indeterminate form, we can apply L'Hôpital's Rule.
03
Apply L'Hôpital's Rule
Use L'Hôpital's Rule, which states if \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \pm \frac{\infty}{\infty} \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \). First, find the derivatives: \( f'(x) = -\sin x \) and \( g'(x) = \cos x \).
04
Recalculate the Limit using Derivatives
The new limit becomes \( \lim_{x \to \pi} \frac{-\sin x}{\cos x} \). When we substitute \( x = \pi \), we get \( \frac{-\sin \pi}{\cos \pi} = \frac{0}{-1} = 0 \). Thus, the limit as \( x \) approaches \( \pi \) is 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit of a Function
The concept of the limit of a function is essential in calculus, as it helps us understand how a function behaves as the input approaches a certain point. When we talk about the limit of a function, we're interested in finding the value that the function is approaching, rather than necessarily the value it takes at that specific point. In mathematical terms, for a function \( f(x) \), the limit as \( x \) approaches a particular value \( a \) can be written as \( \lim_{x \to a} f(x) \). This notation expresses what value \( f(x) \) is getting close to as \( x \) moves nearer to \( a \).
- The concept of limits is crucial in defining other key ideas in calculus, such as derivatives and integrals.
- Limits can help us understand the behavior of functions near points of interest, even if the function is not defined at those points.
Indeterminate Forms
When evaluating limits, you may come across indeterminate forms, which are expressions where direct substitution doesn't provide a clear answer. These forms arise because the behavior of the numerator and the denominator as they approach a point can lead to uncertainty about the overall result. The most common indeterminate form encountered in calculus is \( \frac{0}{0} \). This form suggests that both the numerator and the denominator approach zero, creating a situation where we cannot immediately determine the outcome of the limit.
- L'Hôpital's Rule is often used to resolve these indeterminate forms by taking derivatives of both the numerator and the denominator.
- Other indeterminate forms include \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), and \( \infty - \infty \), which require similar techniques to evaluate their limits.
Trigonometric Functions
Trigonometric functions are fundamental in mathematics, especially when dealing with periodic phenomena and rotations. Functions like \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \) are frequently involved in calculus problems, where understanding their properties is essential.These functions have special characteristics and behave uniquely at certain critical points, often connected to angles like \( \pi \), \( \frac{\pi}{2} \), and \( 2\pi \).
- For instance, \( \sin(\pi) = 0 \) and \( \cos(\pi) = -1 \), which are key values used in calculus problems.
- Knowing these values, we can better manage trigonometric expressions and simplify the evaluation of limits involving trigonometric terms.