Problem 13 - Thomas Calculus in SI Units

Question

Find the limits.( If in doubt, look at the function's graph.) $$\lim _{x \rightarrow 1^{-}} \sin ^{-1} x$$

Step-by-Step Solution

Answer

\( \lim_{x \to 1^{-}} \sin^{-1}(x) = \frac{\pi}{2} \).

Step 1: Understanding the Function

The function provided is the inverse sine function, or arcsine, denoted as \( \sin^{-1}(x) \). The inverse sine function returns the angle whose sine is \(x\). This function is defined for \(-1 \leq x \leq 1\).

Step 2: Recognizing the One-Sided Limit

The problem asks us to find \( \lim_{x \to 1^{-}} \sin^{-1}(x) \). The notation \(1^{-}\) indicates that we need to find the limit as \(x\) approaches 1 from the left side, meaning we are considering values less than 1 that are approaching 1.

Step 3: Evaluating the Limit

For values of \(x\) very close to 1 from the left, \( \sin^{-1}(x) \) approaches \( \sin^{-1}(1) \). Since \( \sin \left( \frac{\pi}{2} \right)=1\), we have \( \sin^{-1}(1) = \frac{\pi}{2} \). Hence, \( \lim_{x \to 1^{-}} \sin^{-1}(x) = \frac{\pi}{2} \).

Step 4: Confirming with the Function's Graph

Looking at the graph of \(y = \sin^{-1}(x)\), it is clear that as \(x\) approaches 1 from the left, the value of \(y\) approaches \( \frac{\pi}{2} \). This matches our previous evaluation.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are intriguing and essential in mathematics because they let us find angles when given trigonometric ratios. The sine inverse function, denoted as \( \sin^{-1}(x) \) or arcsin, is particularly useful. It returns the angle whose sine value is \(x\). This function is defined within a specific domain, specifically \(-1 \leq x \leq 1\), because these are the range limits of the basic sine function.
When using inverse trigonometric functions, it is crucial to remember that their outputs, often referred to as principal values, are also restricted. For \( \sin^{-1}(x) \), the range is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Consider this as the set of possible angles you might obtain, which correspond to the conventional unit circle angles.

• For \(x = -1\), \(\sin^{-1}(x) = -\frac{\pi}{2}\)
• For \(x = 0\), \(\sin^{-1}(x) = 0\)
• For \(x = 1\), \(\sin^{-1}(x) = \frac{\pi}{2}\)

Understanding this function helps in grasping how angles relate to trigonometric ratios, which is especially useful in calculus, physics, and engineering.

One-Sided Limit

The concept of a one-sided limit is a fundamental aspect of calculus. It refers to the value that a function approaches as the input approaches a specific value from one side - either the left or the right. In mathematical notation, this can be written as \( \lim_{x \to a^{+}} f(x) \) or \( \lim_{x \to a^{-}} f(x) \), where \(a^{+}\) indicates approaching from the right, and \(a^{-}\) indicates approaching from the left.
One-sided limits help us understand functions' behavior in more detail, particularly at points where they might be discontinuous or have a different limit from either side. For the given problem, we need to find the left-hand limit \( \lim_{x \to 1^{-}} \sin^{-1}(x) \). This focuses on values just less than 1 approaching 1 from the left.

• One-sided limits help identify asymmetries in functions at specific points.
• They are useful in determining the continuity of a function at a point.
• They can indicate the existence of "jumps" or discontinuous behavior in the function.

Using one-sided limits can ensure comprehensive understanding when analyzing a function's behavior.

Evaluating Limits

Evaluating limits is a cornerstone of calculus, providing insight into a function's behavior as the input approaches a particular point. It involves determining the value that a function "tends to" as the input nears a certain number. This concept helps in understanding continuity, finding derivatives, and solving complex problems involving changes.
In our specific problem, we calculated \( \lim_{x \to 1^{-}} \sin^{-1}(x) \). Here's a step-by-step approach:

• First, understand the function and the range of \( \sin^{-1}(x) \), which is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
• Next, recognize what it means to approach from the left, specifically, considering values like 0.99 or 0.999.
• Finally, see that as these values get closer to 1, the angle approaches \(\frac{\pi}{2}\).

By graphing \(y = \sin^{-1}(x)\) or understanding its behaviors analytically, we are reassured that our calculated limit, \( \frac{\pi}{2} \), matches exactly what the function is doing as \(x\) approaches \(1\) from the left.
Getting to grips with how to properly evaluate limits is vital for mastering calculus concepts, and it becomes an invaluable tool throughout many mathematical applications.