Problem 30
Question
. Let $$ f(x)= \begin{cases}\frac{\ln x}{x-1}, & \text { if } x \neq 1 \\ c, & \text { if } x=1\end{cases} $$ What value of \(c\) makes \(f(x)\) continuous at \(x=1\) ?
Step-by-Step Solution
Verified Answer
The value of \( c \) that makes \( f(x) \) continuous at \( x=1 \) is 1.
1Step 1: Understand the Definition of Continuity
A function is continuous at a point if the limit of the function as it approaches the point from either side is equal to the function's value at that point. For \( f(x) \) to be continuous at \( x = 1 \), \( \lim_{{x \to 1}} f(x) \) must equal \( f(1) = c \).
2Step 2: Calculate the Limit as x Approaches 1
Since \( f(x) \) is defined as \( \frac{\ln x}{x-1} \) for \( x eq 1 \), compute \( \lim_{{x \to 1}} \frac{\ln x}{x-1} \) using L'Hôpital's Rule. This rule applies because both the numerator and denominator approach 0 as \( x \to 1 \). Calculate the derivatives: the derivative of \( \ln x \) is \( \frac{1}{x} \) and of \( x-1 \) is 1, so: \[ \lim_{{x \to 1}} \frac{\ln x}{x-1} = \lim_{{x \to 1}} \frac{\frac{1}{x}}{1} = \lim_{{x \to 1}} \frac{1}{x} = 1. \]
3Step 3: Set the Limit Equal to c and Solve
For continuity, \( c \) must equal \( \lim_{{x \to 1}} f(x) \). From Step 2, this limit is 1. Therefore, set \( c = 1 \).
4Step 4: Verify Continuity
Confirm that with \( c = 1 \), \( f(x) \) indeed becomes a continuous function at \( x = 1 \). The function's limit when approaching 1 from any direction matches the value \( f(1) = 1 \), ensuring continuity.
Key Concepts
Limit of a FunctionL'Hôpital's RuleDiscontinuous PointsCalculus Problem Solving
Limit of a Function
Understanding the limit of a function is crucial in tackling calculus problems, especially those involving continuity. A limit describes the value that a function approaches as the input (or index) approaches some value. In practical terms, when we examine a limit, we're checking how the function behaves as it gets close to a particular point without necessarily reaching it.
To verify the continuity of a function at a point, we must ensure the function's limit from both the left and the right approaches the same value. For instance, in the problem, we need to determine \( \lim_{{x \to 1}} \frac{\ln x}{x-1} \). This means we're interested in the behavior of the function as \( x \) gets closer to \( 1 \) from both sides of the number line.
The concept of limit helps us determine the value function needs to approach at a given point to be continuous, even if it's not well-defined at that point originally.
To verify the continuity of a function at a point, we must ensure the function's limit from both the left and the right approaches the same value. For instance, in the problem, we need to determine \( \lim_{{x \to 1}} \frac{\ln x}{x-1} \). This means we're interested in the behavior of the function as \( x \) gets closer to \( 1 \) from both sides of the number line.
The concept of limit helps us determine the value function needs to approach at a given point to be continuous, even if it's not well-defined at that point originally.
L'Hôpital's Rule
L'Hôpital's Rule is a lifesaver when it comes to finding limits that initially seem indeterminate, usually in the forms \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \). This rule simplifies the process of calculating such limits by allowing us to differentiate the numerator and the denominator separately, and then determining the limit of that new fraction.
In the given exercise, we encounter \( \lim_{{x \to 1}} \frac{\ln x}{x-1} \). Both the numerator \( \ln x \) and the denominator \( x-1 \) approach zero as \( x \to 1 \), making it an ideal candidate for L'Hôpital's Rule. By differentiating \( \ln x \) to \( \frac{1}{x} \) and \( x-1 \) to \( 1 \), we simplify the problem to \( \lim_{{x \to 1}} \frac{1}{x} \), which is much easier to solve.
Applying L'Hôpital's Rule streamlines the process, giving a clear and concise method to resolve limits that initially resist straightforward evaluation.
In the given exercise, we encounter \( \lim_{{x \to 1}} \frac{\ln x}{x-1} \). Both the numerator \( \ln x \) and the denominator \( x-1 \) approach zero as \( x \to 1 \), making it an ideal candidate for L'Hôpital's Rule. By differentiating \( \ln x \) to \( \frac{1}{x} \) and \( x-1 \) to \( 1 \), we simplify the problem to \( \lim_{{x \to 1}} \frac{1}{x} \), which is much easier to solve.
Applying L'Hôpital's Rule streamlines the process, giving a clear and concise method to resolve limits that initially resist straightforward evaluation.
Discontinuous Points
Discontinuous points are spots where a function "jumps," "gaps," or behaves erratically. For a function to be smooth or continuous at a point, the value of the function at the point must equal the limit as it approaches the point. If these aren't equal, the function has a discontinuity, indicating a sudden shift or break.
In our exercise, the function is defined differently at \( x = 1 \) (a potentially discontinuous point) by a constant \( c \) to counteract this. Continuity requires that \( c \) be set to the limit found via L'Hôpital's Rule. By setting \( c \) as this limit, in our case \( 1 \), we can make the function smooth and continuous without visible breaks at \( x = 1 \).
Understanding discontinuous points helps us identify locations where additional conditions or corrections (like defining \( c \)) are necessary to make the function well-behaved across its entire domain.
In our exercise, the function is defined differently at \( x = 1 \) (a potentially discontinuous point) by a constant \( c \) to counteract this. Continuity requires that \( c \) be set to the limit found via L'Hôpital's Rule. By setting \( c \) as this limit, in our case \( 1 \), we can make the function smooth and continuous without visible breaks at \( x = 1 \).
Understanding discontinuous points helps us identify locations where additional conditions or corrections (like defining \( c \)) are necessary to make the function well-behaved across its entire domain.
Calculus Problem Solving
Calculus problem-solving involves a systematic approach to tackle any given problem in the field, requiring both detailed understanding of the principles and the ability to apply these principles effectively.
When solving continuity problems like in the given exercise, it's important to follow structured steps: understand the definition of continuity, calculate necessary limits, apply appropriate rules (like L'Hôpital's), and verify results. This blend of theoretical knowledge and systematic practice ensures solutions are logical and sound.
Moreover, a solid grasp of calculus problem-solving techniques enables mathematicians and students alike to approach even more complex problems with confidence, knowing that each step logically leads to the next.
When solving continuity problems like in the given exercise, it's important to follow structured steps: understand the definition of continuity, calculate necessary limits, apply appropriate rules (like L'Hôpital's), and verify results. This blend of theoretical knowledge and systematic practice ensures solutions are logical and sound.
Moreover, a solid grasp of calculus problem-solving techniques enables mathematicians and students alike to approach even more complex problems with confidence, knowing that each step logically leads to the next.
Other exercises in this chapter
Problem 29
Evaluate each improper integral or show that it diverges. $$ \int_{1}^{e} \frac{d x}{x \ln x} $$
View solution Problem 29
Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow 0}\left(\csc x-\frac{1}{x}\right)$$
View solution Problem 30
Evaluate each improper integral or show that it diverges. $$ \int_{1}^{10} \frac{d x}{x \ln ^{100} x} $$
View solution Problem 30
Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}$$
View solution