Question

Evaluate the following limits or state that they do not exist. $$\lim _{x \rightarrow 1^{-}} \frac{x}{\ln x}$$

Short answer

If so, what is its value? Answer: The limit does not exist.

Step-by-step solution

Analyze the limit from the left

First, let's analyze the limit as x approaches 1 from the left side. This means that we want to look at the behavior of the function as x gets very close to 1, but stays slightly less than 1 (e.g., 0.999, 0.9999, etc.).

Analyze the numerator

As x approaches 1 from the left, the numerator (x) simply approaches 1. There is no issue with the numerator as x approaches 1.

Analyze the denominator

For the denominator, we have the natural logarithm function, \(\ln x\). As x approaches 1 from the left, the natural logarithm will approach 0. It's important to note that the natural logarithm is continuous and defined for x > 0.

Apply L'Hopital's Rule

Since the limit has the form \(\frac{1}{0}\) when x approaches 1, we will apply L'Hôpital's Rule. The L'Hôpital's Rule states that if the limit has the indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can find the limit by taking the derivatives of both the numerator and the denominator, and then computing the limit of that new expression. In our case, we need to take the derivative of both the numerator and the denominator with respect to x. So, we have: $$(\frac{x}{\ln x})' = \frac{(x)'(\ln x) - (\ln x)'x}{(\ln x)^2}$$ We can now compute the derivatives: $$(x)' = 1$$ $$ (\ln x)' = \frac{1}{x}$$ Plugging these into the expression above gives: $$\lim _{x \rightarrow 1^{-}} \frac{1(\ln x) - \frac{1}{x}x}{(\ln x)^2}$$ Simplify this expression: $$\lim _{x \rightarrow 1^{-}} \frac{\ln x - 1}{(\ln x)^2}$$

Evaluate the new limit

Now we need to evaluate the new limit: $$\lim _{x \rightarrow 1^{-}} \frac{\ln x - 1}{(\ln x)^2}$$ As x approaches 1 from the left, \(\ln x\) approaches 0 from the left, so the limit becomes: $$\lim _{x \rightarrow 1^{-}} \frac{0 - 1}{0^2} = \frac{-1}{0}$$

Conclusion

Since the denominator in the limit is approaching 0, the limit does not exist as it will result in a division by 0.

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful technique in calculus for finding limits that initially result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When you encounter these forms, directly calculating the limit might not be possible. Instead, L'Hôpital's Rule allows you to differentiate the numerator and the denominator separately and then take the limit of the resulting expression.

Here are some key points to understand:

• L'Hôpital's Rule can only be applied to certain indeterminate forms. If your limit results in something like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), it might be a good candidate for L'Hôpital's Rule.
• The conditions under which you can apply this rule include that both functions involved should be differentiable near the point of interest, and the original limit must present an indeterminate form.
• Once you've applied the rule by differentiating the numerator and denominator with respect to \(x\), you re-evaluate the limit. Sometimes, you might need to apply the rule more than once.

In our specific problem, the initial form is a \(\frac{1}{0}\) form. Although not one of the standard indeterminate forms, it signals potential for a direct limit analysis, indicating the need to delve deeper into potential changes around the point of interest.

Natural Logarithm

The natural logarithm, denoted as \(\ln x\), is the logarithm to the base \(e\), where \(e\) is approximately 2.71828. This logarithm is fundamental in calculus primarily due to its unique properties related to growth and decay processes.

The function \(\ln x\) is defined for all positive \(x\) values. As \(x\) approaches 1, \(\ln x\) approaches 0. This behavior is crucial when solving limit problems involving logarithms, as seen in our exercise.

• The derivative of \(\ln x\) is \(\frac{1}{x}\). This property is often used when differentiating expressions involving the natural logarithm, especially when employing L'Hôpital's Rule.
• The natural logarithm compares growth rates, and its graph is a gentle curve, starting from negative infinity as \(x\) approaches 0 from the right and rising to infinite heights as \(x\) grows large.
• It's continuous and smooth for all \(x > 0\), meaning there's no break or hole in the graph especially near the point \(x=1\).

In our exercise, as \(x\) approaches 1 from the left, \(\ln x\)'s behavior is central to determining the overall behavior of the limit.

Indeterminate Forms

In calculus, indeterminate forms are expressions that do not initially provide enough information for limit evaluation. Common indeterminate forms include \(\frac{0}{0}\), \(0 \times \infty\), \(\infty - \infty\), \(\frac{\infty}{\infty}\), \(0^0\), \(1^\infty\), and \(\infty^0\).

Understanding indeterminate forms is crucial in calculus for the proper application of tools like L'Hôpital's Rule:

• These forms indicate that direct substitution would yield an expression that doesn't make intuitive sense (e.g., dividing 0 by 0), thus requiring a more nuanced approach.
• L'Hôpital's Rule is frequently applied when dealing with \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) forms. However, if you encounter forms such as \(\frac{1}{0}\), further analysis is necessary as these typically suggest behavior leading to infinity or undefined values.
• In our specific problem, although \(\frac{1}{0}\) is not standardly indeterminate, it hints at exploring the surrounding function behavior more closely, often signaling asymptotic behavior or an unbounded limit.

Understanding and correctly identifying these forms is essential for correctly solving limits, as demonstrated in our exercise where the expression transitions towards a \(\frac{1}{0}\) form, indicating the absence of a finite limit.