Question

Determine the interval(s) on which the following functions are continuous: then analyze the given limits. $$f(x)=\frac{\ln x}{\sin ^{-1} x} ; \lim _{x \rightarrow 1^{-}} f(x)$$

Short answer

Question: Determine the interval(s) of continuity of the given function, \(f(x) = \frac{\ln x}{\sin^{-1} x}\), and find the limit as x approaches 1 from the left side, i.e., \(\lim_{x \rightarrow 1^{-}} f(x)\). Answer: The function \(f(x)\) is continuous on the interval \(0 < x \leq 1\). The limit as x approaches 1 from the left side is 0, i.e., \(\lim_{x \rightarrow 1^{-}} f(x) = 0\).

Step-by-step solution

Identify the domain

Since the function in question is a fraction, we should first determine the domain of both the numerator (natural logarithm) and the denominator (inverse sine). For the numerator, we have \(\ln x\). To have a continuous and valid function, x must be greater than 0 (x > 0). For the denominator, we have \(\sin^{-1} x\). This function is defined and continuous within the range of \([-1, 1]\). Therefore, we have \(-1 \leq x \leq 1\).

Finding the intersection of domains

Now that we have the domain ranges for both the numerator and the denominator, it's time to find their intersection. We combine the two inequalities from Step 1: \(-1 \leq x \leq 1\) (from the denominator) and \(x > 0\) (from the numerator). Taking their intersection, we get: \(0 < x \leq 1\). This is the interval on which \(f(x)\) is continuous.

Analyzing the limit

We need to find \(\lim_{x \rightarrow 1^{-}} f(x)\), which means finding the value of the function \(f(x)\) when approaching 1 from the left side, i.e., x is slightly less than 1. Since the function is continuous on the interval \((0, 1]\), we can plug in the value of \(1\) directly into the function. This gives: $$f(1) = \frac{\ln 1}{\sin^{-1} 1}$$ Now, using basic logarithm and inverse sine properties, we know that: \(\ln 1 = 0\) and \(\sin^{-1} 1 = \frac{\pi}{2}\) Substitute these values back into \(f(1)\): $$f(1) = \frac{0}{\frac{\pi}{2}}$$ Ultimately, we are left with: $$\lim_{x \rightarrow 1^{-}} f(x) = 0$$

Key concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a special type of logarithm that uses the number \( e \) as its base, where \( e \approx 2.718 \). It is a crucial function in many areas of mathematics. For \( \ln x \) to be defined, the input, \( x \), must be a positive real number. This is because the logarithm of zero or a negative number is undefined.
Understanding where \( \ln x \) is valid is important when analyzing the continuity of functions like \( f(x) = \frac{\ln x}{\sin^{-1} x} \). Here, the natural logarithm dictates that \( x > 0 \).

• Some properties of \( \ln x \) include: \( \ln 1 = 0 \) and \( \ln(ab) = \ln a + \ln b \).
• The natural log function is continuous on its domain, which is \((0, \infty)\).
• It’s a monotonically increasing function. This means as \( x \) increases, \( \ln x \) also increases.

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse the process of the trigonometric functions such as sine, cosine, and tangent. The inverse sine, or \( \sin^{-1} x \), is especially significant as it maps the input back to an angle whose sine is the given number.
When working with \( \sin^{-1} x \), it's important to understand its valid input range and output values to ensure continuity.

• The domain of \( \sin^{-1} x \) is \([-1, 1]\) because the sine of any angle can only produce values within this range.
• The output of \( \sin^{-1} x \) is an angle, typically expressed in radians, within \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
• It is continuous over its entire domain, making it reliable for functions involving trigonometric inversions.

Limits

A limit describes the behavior of a function as its input approaches a particular value. Calculating limits is essential for understanding the continuity of functions, especially at boundary points of domains.
In the exercise, we're interested in \( \lim_{x \rightarrow 1^{-}} f(x) \), indicating that we're looking at the behavior of the function as \( x \) approaches 1 from the left, slightly less than 1.

• If a function's limit as \( x \to a \) exists, it tells us that \( f(x) \) gets closer and closer to a certain value as \( x \) approaches \( a \).
• In our case, \( f(x) \) simplifies at \( x = 1 \) to \( \frac{\ln 1}{\sin^{-1} 1} = \frac{0}{\frac{\pi}{2}} = 0 \).
• Understanding limits helps ensure a function is continuous at a point if the limit is equal to the function’s actual value at that point.

Domain of a Function

The domain of a function is a set of inputs for which the function is defined. Determining the domain is crucial for assessing where the function is continuous and valid.
In the exercise, the domain was calculated by determining where both \( \ln x \) and \( \sin^{-1} x \) are defined and then finding their intersection.

• For \( \ln x \), the domain is \((0, \infty)\).
• For \( \sin^{-1} x \), the domain is \([-1, 1]\).
• The domain of the given function \( f(x) = \frac{\ln x}{\sin^{-1} x} \) is therefore the intersection: \( 0 < x \leq 1 \).
• This intersection ensures that the function is continuous where both components are defined.

Understanding the domain helps in analyzing the points where the function can be evaluated without issues.