Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 1} \frac{x-1}{\ln x} $$

Short answer

The limit is 1.

Step-by-step solution

Identifying the Indeterminate Form

First, we plug in the limit value, which is 1, into the function to see if it results in an indeterminate form. For the numerator, when we substitute 1, we get \(1 - 1 = 0\).For the denominator, when \(x\) approaches 1, \(\ln(1) = 0\).Thus, the expression is of the form \(\frac{0}{0}\), which is an indeterminate form.

Applying L'Hôpital's Rule

Since we have an indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's Rule. This rule involves differentiating the numerator and the denominator separately and then taking the limit again.Differentiate the numerator: \(\frac{d}{dx}(x-1) = 1\).Differentiate the denominator: \(\frac{d}{dx}(\ln x) = \frac{1}{x}\).

Evaluating the Limit Using L'Hôpital's Rule

After differentiating, the expression becomes:\[\lim_{x \to 1} \frac{1}{\frac{1}{x}} = \lim_{x \to 1} x\]Now, substitute \(x = 1\) into the new expression:\[\lim_{x \to 1} x = 1\]Thus, the limit evaluates to 1.

Key concepts

Limits

In calculus, the concept of limits is fundamental because it helps us understand the behavior of functions as they approach a particular point. When we discuss limits, we're interested in discovering what value a function gets closer to as the input (or x-value) nears a certain number. In the given exercise, we have to find the limit of the function \( \frac{x-1}{\ln x} \) as \( x \) approaches 1.

Here's a quick guide on dealing with limits:

• Identify if substitution directly gives a clear answer or results in an indeterminate form.
• If substituting the point gives a zero in the denominator or numerator, we need further steps to solve it.
• Use algebraic manipulation or calculus tools like L'Hôpital's Rule to evaluate.

Understanding limits thoroughly can set a strong foundation for deeper calculus topics such as derivatives and integrals. Remember, limits are about the trends and closely approaching values, not necessarily the value a function takes at that exact point.

Indeterminate Forms

An indeterminate form, such as \( \frac{0}{0} \), signals that more work is needed to evaluate a limit. These forms occur often and are a critical component of calculus because they show points where typical algebra might not provide clarity. In our problem, after substituting \( x = 1 \) into \( \frac{x - 1}{\ln x} \), we end with the expression \( \frac{0}{0} \).

Here's what to do in such cases:

• Recognize the indeterminate form: \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), among others, require specific approaches to resolve.
• L'Hôpital's Rule is a valuable tool to simplify and solve these forms.
• Always re-evaluate the new expression after applying the rule to ensure the form has been resolved.

By breaking down the problem into manageable parts, identifying and handling indeterminate forms becomes more approachable. This also helps in solidifying your understanding of limits.

Differentiation

Differentiation is the process of finding the derivative of a function. It essentially tells us the rate of change of a function at any given point. In the context of limits and indeterminate forms, differentiation helps resolve tricky scenarios by simplifying the function's behavior around the point of interest.

Here's how differentiation aids in evaluating limits:

• L'Hôpital's Rule uses differentiation by taking the derivatives of the numerator and denominator independently to simplify the problem.
• In this exercise, we differentiated \( x-1 \) to get 1, and \( \ln x \) to get \( \frac{1}{x} \).
• Once differentiated, the limit becomes straightforward to compute because the indeterminate form is eliminated.

Differentiation thus gives us a mathematical lens to see how functions "move," allowing us to crack open results that first seemed unsolvable due to indeterminate forms.