Question

Limits Evaluate the following limits. Use l'Hópital's Rule when it is comvenient and applicable. $$\lim _{x \rightarrow 1^{-}}\left(\frac{x}{x-1}-\frac{x}{\ln x}\right)$$

Short answer

Question: Determine the limit of the given function as x approaches 1 from the left: $$\lim _{x \rightarrow 1^{-}}\left(\frac{x}{x-1}-\frac{x}{\ln x}\right)$$ Short Answer: The limit of the given function as x approaches 1 from the left is 0.

Step-by-step solution

Check if it's an indeterminate form

Let's plug x=1 into the given function to see if we get an indeterminate form: $$\left(\frac{1}{1-1}-\frac{1}{\ln 1}\right) = \left(\frac{1}{0}-\frac{1}{0}\right)$$ This is an indeterminate form (0/0).

Apply l'Hópital's rule

When applying l'Hópital's Rule, we'll take the derivative of the numerator and denominator separately: $$f(x)=\frac{x}{x-1}-\frac{x}{\ln x}$$ Numerator and Denominator of the function f(x) can be written as: $$N(x)=x(\ln x - (x-1))$$ $$D(x)=(x-1) \ln x$$ Let's find their derivatives: $$N'(x)=\frac{d}{dx}(x(\ln x - (x-1))) = (\ln x - (x-1)) + x(\frac{1}{x}-1) = (\ln x - (x-1)) - (x-1) = \ln x -x+1$$ $$D'(x)=\frac{d}{dx}((x-1) \ln x) = \ln x + (x-1)\frac{1}{x} = \ln x + x -1$$ Now we apply l'Hôpital's rule and rewrite the limit as: $$\lim _{x \rightarrow 1^{-}}\left(\frac{\ln x -x+1}{\ln x + x -1}\right)$$

Check the new limit

Let's plug x=1 into the expression and check if we get a finite value: $$\frac{\ln 1 - 1 + 1}{\ln 1 + 1 - 1} = \frac{0}{0}$$ We still have an indeterminate form, so we need to apply l'Hôpital's rule again.

Apply l'Hôpital's rule again

Let's take the derivative of the new numerator and denominator: $$N'(x)= \frac{d}{dx}(\ln x -x+1) = \frac{1}{x} - 1$$ $$D'(x)= \frac{d}{dx}(\ln x + x - 1) = \frac{1}{x} + 1$$ Applying l'Hôpital's rule again: $$\lim _{x \rightarrow 1^{-}}\left(\frac{\frac{1}{x} - 1}{\frac{1}{x} + 1}\right)$$

Check the limit again

Let's plug x=1 into the expression: $$\frac{\frac{1}{1} - 1}{\frac{1}{1} + 1} = \frac{0}{2}$$ This time, the limit exists and is equal to 0. Therefore, $$\lim _{x \rightarrow 1^{-}}\left(\frac{x}{x-1}-\frac{x}{\ln x}\right) = 0$$

Key concepts

l'Hôpital's Rule

L'Hôpital's Rule is a powerful tool used in calculus to evaluate limits that initially present as indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if you encounter such forms, you can find the limit by taking the derivatives of the numerator and the denominator separately and then re-evaluating the limit.

The process is straightforward but requires both functions in the limit to be differentiable and that the original limit meets the indeterminate form condition. It's worth noting, however, that simply being indeterminate doesn't guarantee l'Hôpital's Rule will give a solution immediately; often, it might need multiple applications. In the given exercise, we had to apply l'Hôpital's Rule twice to resolve the indeterminate form and find the limit. This illustrates how versatile and repeated application of this rule can lead to a solution, especially when evaluating complex limits.

Indeterminate Forms

Indeterminate forms arise in calculus when evaluating limits that produce expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms don't immediately tell us the limit value, which is why they are referred to as indeterminate.

Rather than providing a direct answer, they signify that more analysis is needed using techniques like l'Hôpital's Rule or algebraic manipulation to determine the actual limit.

There are several types of indeterminate forms besides \( \frac{0}{0} \), including \( \infty - \infty \), \( 0 \times \infty \), \( \frac{0}{\pm \infty} \), and more, which each require different approaches. In the exercise, we encountered a \( \frac{0}{0} \) indeterminate form that required l'Hôpital's Rule to simplify and find the correct limit. Identifying the form correctly is crucial for choosing the right method to solve the problem.

Derivative

The derivative is a key concept in calculus that measures how a function changes as its input changes. At a fundamental level, it's the rate of change or slope of the function at a given point.

Mathematically, if \( f(x) \) is a function, its derivative \( f'(x) \) is defined as the limit: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]This formula allows us to calculate the instantaneous rate of change of \( f \) at any point \( x \).

In the context of l'Hôpital's Rule, derivatives make it possible to transform an indeterminate limit into a solvable one by simplifying the expression's numerator and denominator. For the exercise problem, finding the derivatives as required by l'Hôpital's Rule was crucial to evaluate and resolve the limit successfully. Understanding how to differentiate correctly is essential to applying l'Hôpital's Rule effectively.