Question

In Exercises 2.4.2-2.4.40, find the indicated limits. $$ \lim _{x \rightarrow \infty} \frac{x^{x}}{x \log x} $$

Short answer

Answer: The limit of the function \(\frac{x^x}{x \log x}\) as x approaches infinity is \(\infty\).

Step-by-step solution

Rewrite the function in terms of exponentials

To rewrite the function, we'll express \(x^x\) as an exponential function, using the fact that \(a^b = e^{b \ln a}\). So, we have: $$ \frac{x^x}{x \log x} = \frac{e^{x\ln x}}{x \log x} $$

Apply L'Hopital's Rule

Now, we'll apply L'Hopital's Rule to the rewritten function, which involves taking the derivative of the numerator and denominator with respect to x and then finding the limit of the ratio of the derivatives. To apply L'Hopital's rule, we need to check if the limit forms an indeterminate form of type \(\frac{\infty}{\infty}\). Since \(x \rightarrow \infty\), we have \(e^{x\ln x} \rightarrow \infty\) and \(x \log x \rightarrow \infty\): $$ \lim _{x \rightarrow \infty} \frac{e^{x\ln x}}{x \log x} = \lim _{x \rightarrow \infty} \frac{\frac{d}{dx}(e^{x\ln x})}{\frac{d}{dx}(x \log x)} $$

Take the derivatives of numerator and denominator

We need to find the derivative of the numerator \(e^{x\ln x}\) and denominator \(x \log x\): The derivative of the numerator using Chain Rule: $$ \frac{d}{dx}(e^{x\ln x}) = e^{x\ln x} \frac{d}{dx}(x\ln x) = e^{x\ln x} (1 + \ln x) $$ The derivative of the denominator using Product Rule: $$ \frac{d}{dx}(x \log x) = x \left(\frac{1}{x \ln 10}\right) + \log x = 1 + \frac{\log x}{\ln 10} $$ Now substitute the derivatives back into the L'Hopital's Rule formula: $$ \lim _{x \rightarrow \infty} \frac{e^{x\ln x}(1 + \ln x)}{1 + \frac{\log x}{\ln 10}} $$

Simplify and find the limit

Divide both numerator and denominator by \(x\) to simplify the expression: $$ \lim _{x \rightarrow \infty} \frac{e^{x\ln x}(1 + \ln x)}{1 + \frac{\log x}{\ln 10}} = \lim _{x \rightarrow \infty} \frac{e^{x\ln x}\frac{(1 + \ln x)}{x}}{\frac{1}{x} + \frac{\log x}{x\ln 10}} $$ Take the limit as \(x \rightarrow \infty\): $$ \lim _{x \rightarrow \infty} \frac{e^{x\ln x}\frac{(1 + \ln x)}{x}}{\frac{1}{x} + \frac{\log x}{x\ln 10}} = \frac{\infty}{0} = \infty $$ So, the limit of the given function as x approaches infinity is: $$ \lim _{x \rightarrow \infty} \frac{x^{x}}{x \log x} = \infty $$

Key concepts

L'Hopital's Rule

L'Hopital's Rule is a powerful technique in calculus used to evaluate limits that result in indeterminate forms, such as \(0/0\) or \(\infty/\infty\). When confronted with a limit of a ratio of two functions that yields an indeterminate form, L'Hopital's Rule states that the limit of the ratio is equivalent to the limit of the ratios of their derivatives, provided that certain conditions are met.

The rule requires that both the numerator and denominator be differentiable near the point of interest and that their limit individually approaches zero or infinity. When applying the rule, it's common to continue to differentiate the numerator and denominator until a determinate form is attained, at which point the limit can be calculated.

For example, in our exercise, after rewriting the function \(\lim _{x \rightarrow \infty} \frac{x^{x}}{x \log x}\) as \(\lim _{x \rightarrow \infty} \frac{e^{x\ln x}}{x \log x}\), we recognize an \(\infty/\infty\) form. L'Hopital's Rule is applied by differentiating the numerator \(e^{x\ln x}\) and the denominator \(x \log x\) separately and then taking the limit of the new ratio.

Indeterminate Forms

Indeterminate forms arise when evaluating limits in calculus and represent expressions whose behavior is not immediately obvious. These include \(0/0\), \(0\cdot\infty\), \(\infty/\infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), and \(\infty^0\). They are called 'indeterminate' because the limit could potentially be any value or even nonexistent, depending on the behavior of the functions involved.

In the case of our problem, as \(x\) approaches infinity, both the numerator \(e^{x\ln x}\) and the denominator \(x \log x\) go to infinity. This creates an \(\infty/\infty\) indeterminate form. It's important to resolve such forms to determine the specific behavior of the limit, and L'Hopital's Rule is one of the methods used to clarify the outcome by translating the problem into something more manageable using derivatives.

Derivative Calculus

Derivative calculus, at its core, involves finding the rate at which a function changes at any given point. Derivatives are fundamental for solving problems in calculus, particularly when dealing with limits, as seen in L'Hopital's Rule. The derivative tells us the slope of the tangent line to the curve of the function at any point, essentially giving us the instant rate of change.

In our exercise, the derivatives of \(e^{x\ln x}\) and \(x \log x\) needed careful attention. The differentiation of \(e^{x\ln x}\) employed the Chain Rule, resulting in \(e^{x\ln x} (1 + \ln x)\), while the Product Rule was used for \(x \log x\), yielding \(1 + \frac{\log x}{\ln 10}\). Understanding how to properly apply differentiation rules is crucial to simplifying expressions and ultimately determining the limits of functions.

These derivatives replaced the original functions within the context of L'Hopital's Rule, leading us to find the limit of their ratio as \(x\) approaches infinity. Derivative calculus turned the indeterminate form into a ratio that could be further simplified to find that the limit of \(\frac{x^{x}}{x \log x}\) as \(x\) approaches infinity is, in fact, infinity.