Question

Let $$f(x)=\left\\{\begin{array}{ll} \frac{\ln x}{x-1}, & \text { if } x \neq 1 \\ c, & \text { if } x=1 \end{array}\right. $$ What value of \(c\) makes \(f(x)\) continuous at \(x=1\) ?

Short answer

The value of \( c \) that makes \( f(x) \) continuous at \( x = 1 \) is 1.

Step-by-step solution

Define Continuity

For the function \( f(x) \) to be continuous at \( x = 1 \), the following must be true: \( f(1) = \lim_{{x \to 1}} f(x) \). This means the function value at \( x = 1 \) must equal the limit of \( f(x) \) as \( x \) approaches 1.

Calculate the Limit

Calculate \( \lim_{{x \to 1}} \frac{\ln x}{x-1} \). This limit results in an indeterminate form \( \frac{0}{0} \), so we apply L'Hôpital's Rule.

Apply L'Hôpital's Rule

L'Hôpital's Rule states that \( \lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)} \) if \( \lim_{{x \to c}} f(x) = 0 \) and \( \lim_{{x \to c}} g(x) = 0 \) or both are infinite. Here, differentiate the numerator and the denominator: - \( f'(x) = (\ln x)' = \frac{1}{x} \) - \( g'(x) = (x-1)' = 1 \) Apply the rule: \[ \lim_{{x \to 1}} \frac{\ln x}{x-1} = \lim_{{x \to 1}} \frac{\frac{1}{x}}{1} = \lim_{{x \to 1}} \frac{1}{x} = \frac{1}{1} = 1 \]

Determine the Value of \(c\)

Since \( f(x) \) is continuous at \( x = 1 \) only if \( f(1) = \lim_{{x \to 1}} f(x) \), set \( c = \lim_{{x \to 1}} \frac{\ln x}{x-1} = 1 \).

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool used in calculus to find limits that initially result in indeterminate forms of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule allows us to simplify the limit calculation by differentiating the function's numerator and denominator. When applying L'Hôpital's Rule, we essentially replace the original functions with their derivatives and then re-evaluate the limit.

To apply this rule correctly, it's crucial to ensure both the numerator and denominator approach zero or infinity. Only then can we take the derivatives and thus simplify our problem significantly. Once we find those derivatives, we take the limit of their quotient. This often leads to a much simpler expression, allowing easier limit calculation.

Indeterminate Form

An indeterminate form arises in calculus when the direct substitution in a limit results in an undefined expression. Common indeterminate forms include \( \frac{0}{0}, \frac{\infty}{\infty}, \infty - \infty \), or \( 0 \times \infty \). Each of these forms requires a specific approach to resolve.

In our problem, directly substituting \( x = 1 \) into the function \( \frac{\ln x}{x - 1} \) leads to \( \frac{0}{0} \), an indeterminate form. This signifies that neither the numerator nor the denominator conclusively determines the limit's behavior. Therefore, techniques such as factoring, simplifying, or applying L'Hôpital's Rule become necessary. These strategies help transform the expression into a determinate form that can be easily evaluated.

Limit Calculation

Limit calculation is an essential concept in calculus, needed to understand how functions behave near specific points. Often, direct substitution isn't enough, particularly if it results in indeterminate forms. Instead, limits help us examine the behavior of a function as it approaches a particular value.

In the exercise, solving the limit \( \lim_{{x \to 1}} \frac{\ln x}{x - 1} \) presented a challenge due to its indeterminate form. Using L'Hôpital's Rule, we differentiate the numerator and denominator, yielding \( \frac{1/x}{1} = \frac{1}{x} \). Evaluating this as \( x \to 1 \), we find the limit is equal to 1. This limit calculation is vital for determining how the function behaves at the specified point.

Function Value

The function value, particularly at a given point, is the number that the function output results in when the input is inserted into its formula. For continuity, the function value at this point must match the limit of the function as it approaches that point.

In our problem, the function is defined as \( f(x) = \frac{\ln x}{x-1} \) for \( x eq 1 \) and \( f(x) = c \) for \( x = 1 \). To ensure the function is continuous at \( x = 1 \), \( c \) must equate to the limit as \( x \) approaches 1. Given our limit calculation produced a result of 1, setting \( c = 1 \) makes \( f(x) \) continuous at \( x = 1 \). This means the function does not break or jump at this point, ensuring a smooth continuous graph.