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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.
To find the limit, it sometimes requires more than plugging numbers into a function. This is particularly true when the direct substitution leads to indeterminate forms, requiring the use of advanced techniques like L'Hôpital's Rule.
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.
Understanding and identifying these forms is crucial for accurately evaluating limits and ensuring that results are mathematically sound.
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.
By thoroughly understanding trigonometric functions, you can solve problems more efficiently, particularly when these functions are involved in complex calculus operations like finding limits or derivatives.

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