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Chapter 2: Problem 43

Find the limits. $$\lim _{x \rightarrow 0}(2 \sin x-1)$$

Short Answer

Expert verified
The limit is -1.

Step by step solution

01

Understand the Expression

The function given is \( f(x) = 2 \sin x - 1 \). We need to find the limit of this function as \( x \) approaches 0.
02

Evaluate the Sine Function at 0

Recall that \( \sin x \) is a continuous function and \( \sin(0) = 0 \). This means we can substitute \( x = 0 \) directly into the sine function.
03

Calculate the Value of the Given Function at x = 0

Substitute \( \sin(0) = 0 \) back into the expression: \( 2\sin(0) - 1 = 2 \times 0 - 1 = -1 \).
04

Conclude the Limit

Since \( \sin x \) is continuous and we have evaluated it directly at \( x = 0 \), we conclude that the limit \( \lim_{x \to 0}(2 \sin x - 1) = -1 \).

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Key Concepts

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

Sine Function
The sine function, denoted as \( \sin x \), is one of the fundamental functions in trigonometry. It describes a smooth, repeating wave pattern, which is central to many mathematical and real-world applications, such as sound waves and alternating current.
  • *Wave Nature*: The sine function maps angles to their sine values, giving outputs that range from -1 to 1.
  • *Periodicity*: Sine waves repeat every \( 2\pi \), meaning \( \sin(x + 2\pi) = \sin x \).
  • *Symmetry*: The function is odd, so \( \sin(-x) = -\sin x \).
When dealing with calculus and limits, the sine function is crucial due to its well-understood behavior and smoothness.
Continuous Functions
Continuous functions are those that exhibit no sudden jumps or interruptions in their graphs. For a function to be continuous at a point, the following must hold:
  • *The function is defined at that point.*
  • *The limit of the function exists as it approaches the point.*
  • *The function's value equals its limit at that point.*
The sine function is a classic example of a continuous function. Since it is continuous everywhere, including at \( x = 0 \), limits involving sine functions can often be evaluated through direct substitution.
This property was used in the step-by-step solution to plug \( x = 0 \) into \( \sin x \) and find the limit of the overall function easily.
Evaluating Limits
Evaluating limits is a foundational concept in calculus. It involves finding the value that a function approaches as the input approaches a certain point.

There are several ways to evaluate limits, such as:
  • **Direct Substitution**: Simply substitute the point into the function, which works well when dealing with continuous functions, like in this problem.
  • **Factoring and Simplifying**: Useful for functions that initially appear to be undefined at the point.
  • **L'Hopital's Rule**: Applies when direct substitution leads to indeterminate forms like \( \frac{0}{0} \).
In the provided problem, direct substitution was used. Knowing that \( \sin x \) is continuous, we directly evaluated the function at \( x = 0 \) to find \( \lim_{x \to 0}(2 \sin x - 1) = -1 \), demonstrating that this method can straightaway give us the solution when applicable.

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Most popular questions from this chapter

If the product function \(h(x)=f(x) \cdot g(x)\) is continuous at \(x=0\) must \(f(x)\) and \(g(x)\) be continuous at \(x=0 ?\) Give reasons for your answer.

Find the limits \(\lim \left(\frac{1}{x^{1 / 3}}-\frac{1}{(x-1)^{4 / 3}}\right)\) as a. \(x \rightarrow 0^{+}\) b. \(x \rightarrow 0\) c. \(x \rightarrow 1^{+}\) d. \(x \rightarrow 1^{-}\)

Suppose that \(f(x)\) and \(g(x)\) are polynomials in \(x .\) Can the graph of \(f(x) / g(x)\) have an asymptote if \(g(x)\) is never zero? Give reasons for your answer.

Find the limits $$\lim _{x \rightarrow-\infty}(\sqrt{x^{2}+3}+x)$$

Graph the curves. Explain the relationship between the curve's formula and what you see. $$y=\frac{-1}{\sqrt{4-x^{2}}}$$
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