Problem 54 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Find the domain of \(\ln{x}\)

The natural logarithm function, \(\ln{x}\), is defined and continuous for \(x>0\). This means the domain of \(\ln{x}\) is \((0,\infty)\).

Step 2: Find the domain of \(\sin^{-1}{x}\)

The inverse sine function, \(\sin^{-1}{x}\), is defined for all \(x\) in the closed interval \([-1,1]\). This means the domain of \(\sin^{-1}{x}\) is \([-1,1]\).

Step 3: Find the intersection of both domains

To find the domain of the given function, we need to find the intersection of the domains of both \(\ln{x}\) and \(\sin^{-1}{x}\). The intersection interval is \((0, 1]\). Hence, this is the interval on which the given function is continuous.

Step 4: Analyze the given limit

We need to find the limit: $$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} \frac{\ln{x}}{\sin^{-1}{x}}$$ The limit is taken as \(x\) approaches \(1\) from the left side. Since the function is continuous on the interval \((0,1]\), we can plug the value \(x=1\) directly into the function to find the limit:

Step 5: Substitute the value into the function

Substituting \(x=1\) into the function, we get: $$f(1)=\frac{\ln{1}}{\sin^{-1}{1}}$$ We know that \(\ln{1}=0\) and \(\sin^{-1}{1}=\frac{\pi}{2}\). Therefore,

Step 6: Calculate the limit

The limit of the function as \(x\) approaches \(1\) from the left is: $$\lim_{x \rightarrow 1^{-}} f(x) = \frac{0}{\frac{\pi}{2}} = 0$$ In conclusion, the given function is continuous on the interval \((0, 1]\), and the limit as \(x\) approaches \(1\) from the left side is \(0\).

Key Concepts

Domain of Logarithmic Functions

Understanding the domain of logarithmic functions is an essential step in analyzing continuity and limits. Logarithms are only defined for positive real numbers, which stems from the way logarithmic functions invert exponential functions. For the natural logarithm, denoted as \( \ln(x) \), the domain strictly includes all positive real numbers — in mathematical terms, this is \( (0, \infty) \).

The constraints on the domain exist because there are no real number exponents that can raise a positive base like the number \( e \) (approximately 2.71828) to a negative or zero result. Hence, when dealing with logarithmic functions, you should always remember that the argument of the log function must be positive. In the context of homework exercises, first identify the domain to ensure the values you work with are valid inputs for the logarithm.

Domain of Inverse Trigonometric Functions

Inverse trigonometric functions, like the arcsine function, written as \( \sin^{-1}(x) \), have specific domains that restrict the values they can operate on. The domain for \( \sin^{-1}(x) \) is \( [-1, 1] \) because the sine of an angle in a right triangle cannot exceed 1 nor be less than -1, according to the definition of sine in terms of the side lengths of a right triangle.

Consequently, when carrying out problems that involve these functions, checking that the input values fall within the domain is crucial for a correct solution. For example, if you encounter an expression such as \( \sin^{-1}(x) \) in a problem, you would immediately know that the value of \( x \) must be between -1 and 1 inclusive, else the expression would not be valid.

Limits of Functions

The concept of limits is foundational in calculus and helps us understand the behavior of functions as they approach a certain point. It forms the basis of defining continuity and differentiability. When you come across \( \lim_{x \rightarrow c} f(x) \), you're being asked what value \( f(x) \) approaches as \( x \) gets increasingly close to \( c \).

It's important to note that the limit does not necessarily have to equal the value of the function at that point — in fact, a function can have a limit at a point even if it's undefined there. The limit primarily describes the trend of the function's value as \( x \) approaches from either the left (denoted \( x \rightarrow c^- \) ) or the right (denoted \( x \rightarrow c^+ \) ). When evaluating limits, especially those approaching a point where the function changes nature or the domain ends, one often needs to consider one-sided limits to capture the correct behavior of the function.

Interval of Continuity

The interval of continuity for a function describes the range over which the function is continuous. A function is continuous at a point if a small change in \( x \) results in a small change in \( f(x) \) and the function is defined there. We say a function is continuous on an interval if it is continuous at every point within that interval.

The intersection of domains of all constituent parts of a function determines its interval of continuity. In the given exercise, the continuity of \( f(x) = \frac{\ln x}{\sin ^{-1} x} \) was determined by identifying the intersection of the domains of \( \ln x \) (which is \( (0,\infty) \) ) and \( \sin^{-1} x \) (which is \( [-1,1] \) ). The only values that fit both conditions are those in the range \( (0, 1] \) – hence, this forms the interval of continuity for \( f(x) \).