Using L`HOSPITAL Rule, find Limit x tends to zero (logx)^(log(x-1))lim(x→0) log

To find the limit of \((\log x)^{\log(x-1)}\) as \(x\) approaches zero, we need to analyze the behavior of both \(\log x\) and \(\log(x-1)\) in this context. This limit can be tricky due to the logarithmic functions involved, especially since \(\log x\) approaches negative infinity as \(x\) approaches zero from the right. Let's break this down step by step.

Understanding the Components

First, let's look at the two components of our expression:

• \(\log x\): As \(x\) approaches zero from the right, \(\log x\) tends to \(-\infty\).
• \(\log(x-1)\): As \(x\) approaches zero, \(x-1\) approaches \(-1\), and thus \(\log(x-1)\) is not defined for \(x\) in this range. However, we can consider the limit as \(x\) approaches 1 from the left, where \(\log(x-1)\) approaches \(-\infty\).

Rewriting the Expression

To simplify our limit, we can rewrite the expression using the properties of logarithms. We can express the limit in exponential form:

Let \(y = (\log x)^{\log(x-1)}\). Taking the natural logarithm of both sides gives us:

\(\log y = \log(x-1) \cdot \log(\log x)\)

Finding the Limit of \(\log y\)

Now, we need to evaluate the limit of \(\log y\) as \(x\) approaches zero:

\(\lim_{x \to 0} \log y = \lim_{x \to 0} \log(x-1) \cdot \log(\log x)\)

Analyzing Each Component

As \(x\) approaches zero:

However, we can analyze the limit more carefully. As \(x\) approaches zero, \(\log(\log x)\) approaches \(\log(-\infty)\), which is also undefined. Thus, we need to consider the product of these two terms.

Applying L'Hôpital's Rule

To apply L'Hôpital's Rule, we can rewrite our limit in a form suitable for the rule. We can express the limit as:

\(\lim_{x \to 0} \frac{\log(\log x)}{\frac{1}{\log(x-1)}}\)

Now, both the numerator and denominator approach \(-\infty\) as \(x\) approaches zero. We can apply L'Hôpital's Rule, which states that if we have an indeterminate form of type \(-\infty / -\infty\), we can differentiate the numerator and denominator:

Calculating Derivatives

The derivative of the numerator \(\log(\log x)\) is:

\(\frac{1}{\log x} \cdot \frac{1}{x}\)

The derivative of the denominator \(\frac{1}{\log(x-1)}\) is:

\(-\frac{1}{(\log(x-1))^2} \cdot \frac{1}{x-1}\)

Evaluating the Limit

Now we can substitute these derivatives back into our limit:

\(\lim_{x \to 0} \frac{\frac{1}{\log x} \cdot \frac{1}{x}}{-\frac{1}{(\log(x-1))^2} \cdot \frac{1}{x-1}}\)

As \(x\) approaches zero, both \(\log x\) and \(\log(x-1)\) approach \(-\infty\), leading to a more manageable limit. After simplifying, we find that the limit approaches zero.

Final Result

Thus, we conclude that:

\(\lim_{x \to 0} (\log x)^{\log(x-1)} = e^{\lim_{x \to 0} \log y} = e^0 = 1\).

In summary, the limit of \((\log x)^{\log(x-1)}\) as \(x\) approaches zero is equal to 1. This process illustrates the importance of understanding logarithmic behavior and applying L'Hôpital's Rule effectively to resolve indeterminate forms.