Problem 41 - Thomas Calculus

Question

Find the limits in Exercises \(41-48 .\) (If in doubt, look at the function's graph.) $$ \lim _{x \rightarrow 1^{-}} \sin ^{-1} x $$

Step-by-Step Solution

Answer

The limit is \( \frac{\pi}{2} \).

Step 1: Understand the notation

The notation \( \lim _{x \rightarrow 1^{-}} \sin ^{-1} x \) represents the left-hand limit of the function \( \sin^{-1} x \) as \( x \) approaches 1 from the left.

Step 2: Know the domain and range of arcsin

The inverse sine function, \( \sin^{-1} x \), is defined for \(-1 \leq x \leq 1\) and outputs values in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). As \( x \rightarrow 1 \), the function approaches its maximum value.

Step 3: Substitute the value into the function

Substitute a value very close to but less than 1, such as \( x = 0.999 \), into \( \sin^{-1} x \). This gives \( \sin^{-1}(0.999) \approx \frac{\pi}{2} \) rad.

Step 4: Analyze the behavior

As \( x \rightarrow 1^{-} \), \( \sin^{-1} x \) approaches \( \frac{\pi}{2} \) continually. This confirms that the left-hand limit of \( \sin^{-1} x \) as \( x \) approaches 1 is consistent with substituting values close to 1.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the functions that reverse the action of the usual trigonometric functions, like sine, cosine, and tangent. For example, arcsin or \( \sin^{-1} \) is the inverse of the sine function. This means if \( \sin(y) = x \), then \( y = \sin^{-1}(x) \). Here are some key things to know:

• Domain and Range: The domain of \( \sin^{-1}(x) \) is \(-1 \leq x \leq 1\). This is because the output of the sine function also falls within this interval.
• Output: The range of \( \sin^{-1}(x) \) is \(-\frac{\pi}{2} \leq \sin^{-1}(x) \leq \frac{\pi}{2}\), which means the inverse function will provide angles in radians that fall between these values.
• Applications: It's widely used in trigonometry to find angles when the sine value is known, and this helps in solving different types of equations in calculus.

Inverse trigonometric functions make it easier to handle problems related to periodic nature, angles, and rotations in calculus and geometry.

Left-Hand Limits

Left-hand limits are one-sided limits where we approach a point from the left-hand side on a graph. In mathematical notation, it is expressed as \( \lim_{x \rightarrow a^{-}} f(x) \). Understanding left-hand limits is crucial for analyzing how a function behaves as it gets close to a certain value:

• It deals with values that are less than the point of interest. For example, in \( \lim_{x \rightarrow 1^{-}} \sin^{-1} x \), \( x \) approaches 1 but remains slightly less than 1.
• Left-hand limits help determine a function's behavior near discontinuities or points of interest, where its value might suddenly change.
• Unlike full limits, which consider both sides, left-hand limits only use one approach direction, giving specific insights into function behavior.

By honing in on one side, left-hand limits aid in better understanding and visualizing functions and their breaking points on a graph.

Continuous Functions

Continuous functions are functions with no breaks, jumps, or holes in their graphs. They are predictable and can be traced without lifting a pencil from the paper:

• A function \( f \) is continuous at a point \( x = a \) if the left-hand limit, right-hand limit, and the function's value at \( a \) are all equal: \( \lim_{x \rightarrow a} f(x) = f(a) \).
• Inverse trigonometric functions like \( \sin^{-1}(x) \) are continuous over their whole domain because they possess no breaks within the permitted range of the angle.
• Continuous functions ensure a smooth transition across their domain, which makes them crucial for studies involving limits, derivatives, and integrals in calculus.

The concept of continuity is foundational in calculus, as it helps in determining the viability of applying certain mathematical operations or concepts, especially when analyzing more advanced problems.