/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Find the limits. $$\lim _{x \r... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the limits. $$\lim _{x \rightarrow-\pi / 6} \sqrt{1+\cos (\pi \csc x)}$$

Short Answer

Expert verified
The limit is \(\sqrt{2}\).

Step by step solution

01

Analyze the Limit Expression

We need to find the limit as \(x\) approaches \(-\pi/6\) for the expression \(\sqrt{1 + \cos(\pi \csc x)}\). The expression involves trigonometric functions, so we should first consider the behavior of those functions as \( x \to -\pi/6 \).
02

Evaluate \(\csc x\) at the Limit

As \(x \to -\pi/6\), \(\csc x = 1/\sin x\). At \(x = -\pi/6\), \(\sin(-\pi/6) = -1/2\), so \(\csc(-\pi/6) = -2\). Thus, \(\pi \csc x \to \pi(-2) = -2\pi\) as \(x \to -\pi/6\).
03

Evaluate \(\cos(\pi \csc x)\) at the Limit

Substitute the limit we found for \(\pi \csc x\) into the cosine function, \(\cos(-2\pi)\). Since cosine is periodic with period \(2\pi\), \(\cos(-2\pi) = \cos(0) = 1\).
04

Simplify the Expression

Now substitute this result back into the original expression: \(\sqrt{1 + \cos(-2\pi)} = \sqrt{1 + 1} = \sqrt{2}\).
05

Conclude the Limit

The original limit expression simplifies to \(\sqrt{2}\) as \(x\) approaches \(-\pi/6\). Therefore, the limit is \(\sqrt{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Trigonometric functions relate the angles of a triangle to the lengths of its sides and play a crucial role in mathematics. The most common trigonometric functions are sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)). They are periodic, meaning their values repeat at regular intervals. For instance, the sine and cosine functions have a period of \( 2\pi \). This means that \( \sin(x) = \sin(x + 2\pi) \) and \( \cos(x) = \cos(x + 2\pi) \) for all values of \( x \).
When working with trigonometric functions in limits, it is important to understand their behavior at specific angles. The function \( \csc(x) \) is the reciprocal of sine: \( \csc(x) = 1/\sin(x) \). For example, \( \sin(-\pi/6) = -1/2 \), so \( \csc(-\pi/6) = -2 \). Understanding these relationships lets us evaluate expressions involving trigonometric functions as \( x \) approaches a particular value.
Trigonometric functions are ubiquitous in calculus problems, especially those involving limits and continuity. They often require manipulation and substitution to simplify expressions and evaluate limits successfully.
Limit Evaluation
Limits are a fundamental concept in calculus used to determine the value that a function approaches as the input approaches a particular point. In the context of the exercise, we are evaluating the limit \( \lim_{x \rightarrow -\pi/6} \sqrt{1 + \cos(\pi \csc x)} \).
Evaluating limits often requires understanding how expressions behave around the point of interest and using algebraic manipulations to simplify them. In our problem:
  • First, we calculate \( \csc x \) at \( x = -\pi/6 \) to find \( \pi \csc x \).
  • Next, we evaluate \( \cos(\pi \csc x) \), recognizing that cosine is periodic, and simplify to find \( \cos(-2\pi) = 1 \).
Using these calculations, we substitute back into the original limit expression. Finally, solve the limit expression \( \sqrt{1 + 1} \), which yields the result \( \sqrt{2} \).
This process illustrates how limit evaluation requires careful manipulation of functions and analysis of their behavior around specific points.
Continuity Analysis
In mathematics, a function is said to be continuous on an interval if it is defined at every point of the interval, and small changes in input result in small changes in output. Continuity helps ascertain the behavior of functions, especially while evaluating limits.
The exercise entails evaluating an expression involving \( \cos(x) \), a continuous function over all real numbers. Because the cosine function is continuous, evaluating limits of expressions involving cosine often reduces to simple substitution once we find input values.
Continuity also implies that small variations in \( x \) near a known value like \( -\pi/6 \) will not cause any jumps in the value of \( \cos(\pi \csc x) \). This stability simplifies the limit evaluation process, turning a potentially complex problem into a straightforward calculation.
Overall, continuity analysis is essential in understanding how functions behave under limit evaluations, ensuring that their innate smoothness makes solving calculus problems simpler and more intuitive.

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