Problem 43 - Thomas Calculus

Question

Limits with trigonometric functions Find the limits in Exercises \(43-50 .\) $$\lim _{x \rightarrow 0}(2 \sin x-1)$$

Step-by-Step Solution

Answer

The limit is -1.

Step 1: Understand the Problem

We need to find the limit of the function \( 2 \sin x - 1 \) as \( x \) approaches 0.

Step 2: Substitute Directly

Since the function is continuous, substitute \( x = 0 \) directly into the function. Compute \( 2 \sin(0) - 1 \).

Step 3: Calculate \( \sin(0) \)

Recall that the sine of 0 radians is 0. Therefore, \( \sin(0) = 0 \).

Step 4: Simplify the Expression

Replace \( \sin(0) \) in the expression \( 2 \sin(0) - 1 \) with 0, yielding \( 2(0) - 1 = 0 - 1 = -1 \).

Step 5: Conclude the Limits

The limit of \( 2 \sin x - 1 \) as \( x \rightarrow 0 \) is \( -1 \).

Key Concepts

Continuous Functions

In mathematics, continuous functions are key to understanding limits and calculus overall. A function is considered continuous if there are no breaks, jumps, or holes in its graph. This means you can draw it without lifting your pencil from the paper. For a function to be continuous at a point, the following must hold:

• The function is defined at the point.
• The limit exists at that point.
• The limit of the function as it approaches the point from both sides equals the value of the function at that point.

In the given exercise, the function to analyze is continuous everywhere because it is composed of basic operations (addition, subtraction, multiplication) on the sine function, which is continuously defined for all real numbers. This inherent continuity allows us to directly substitute the value of the point into the function when calculating limits.

Sine Function

The sine function is an essential trigonometric function that appears frequently in calculus. It is defined for all real numbers and has a range of values between -1 and 1. The formula for the sine of an angle in radians, particularly for small angles close to 0, can often be approximated as:\[ \sin(x) \approx x \]for values of \( x \) close to zero. However, when calculating the exact sine of 0, as required in the exercise, it is precisely 0. This property simplifies many trigonometric limit problems, allowing us to evaluate expressions straightforwardly. Additionally, the sine function's graph is periodic, meaning it repeats its shape over regular intervals of \( 2\pi \), further contributing to its unique characteristics and usefulness in solving trigonometric limits.

Limit Calculation

Evaluating limits is an integral part of calculus as it helps understand the behavior of functions at specific points, often where they cannot be explicitly calculated. In general, the limit of a function \( f(x) \) as \( x \) approaches a value \( c \) denotes the value that \( f(x) \) gets closer to as \( x \) gets closer to \( c \). In the original step-by-step exercise:1. Direct substitution was used due to the continuity of the involved trigonometric function.2. We substituted \( x = 0 \) into the expression \( 2 \sin x - 1 \), relying on the continuity of the sine function.3. The correct evaluation of \( \sin(0) = 0 \) allowed simplification of the limit expression to find the final value.Limits are foundational in understanding derivatives and integrals, and mastering their calculation is key to progressing in calculus.