Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 0}\left(x \ln x^{1000}\right) $$

Short answer

The limit is 0.

Step-by-step solution

Rewrite the Expression

Initially, the limit expression is given as \( \lim_{x \to 0}(x \ln x^{1000}) \). We can simplify this expression by using the logarithmic property \( \ln(a^b) = b \ln a \). Thus, we can rewrite the expression as \( x \cdot 1000 \ln x \). This simplifies to \( 1000x \ln x \).

Check for Indeterminate Form

We need to determine the form of \( 1000x \ln x \) as \( x \to 0 \). As \( x \to 0^+ \), \( \ln x \to -\infty \) and \( x \to 0 \). Therefore, the product \( 1000x \ln x \) takes the form \( 0 \times (-\infty) \), which is an indeterminate form. To apply l'Hôpital's Rule, we should convert this into a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form.

Convert to a Fraction

To convert \( 1000 x \ln x \) into a form suitable for l'Hôpital's Rule, rewrite it as \( \frac{\ln x}{1/x} \). This is a \( \frac{\infty}{\infty} \) form as \( x \to 0^+ \), because \( \ln x \to -\infty \) and \( \frac{1}{x} \to \infty \). Now, we can apply l'Hôpital's Rule.

Apply l'Hôpital's Rule

By applying l'Hôpital's Rule, which involves differentiating the numerator and the denominator separately, we differentiate \( \ln x \) to get \( \frac{1}{x} \) and \( \frac{1}{x} \) to get \( -\frac{1}{x^2} \). Thus, the limit becomes \( \lim_{x \rightarrow 0^+} \frac{1/x}{-1/x^2} = \lim_{x \rightarrow 0^+} -x \).

Evaluate the Limit

Now, find \( \lim_{x \rightarrow 0^+} -x \). As \( x \to 0^+ \), the expression \(-x\) approaches \( 0 \). Therefore, the limit is 0.

Key concepts

Indeterminate Forms

When we deal with limits, we often encounter expressions that don't have an immediate or obvious solution. These are called indeterminate forms. An indeterminate form occurs when substitution directly into a limit expression results in non-definitive values such as \( \frac{0}{0} \), \( \infty - \infty \), \( 0 \cdot \infty \), and others. In the given problem, as \( x \to 0^+ \), the term \( x \ln x^{1000} \) simplifies to \( 1000x \ln x \). Here, \( x \to 0 \) suggesting \( 0 \cdot -\infty \), which is indeed one of the classic indeterminate forms. When you face such expressions, it's essential to first confirm that it is in an indeterminate form before applying appropriate techniques like l'Hôpital's Rule to resolve them. This verification step ensures that the approach used is valid.

Limits

Limits are a fundamental concept in calculus that represent the behavior of a function as the input (or variable) approaches a particular value. In practical terms, limits help us understand the value that a function is approaching, even if it might not actually reach that value for any finite input.In the exercise, we are asked to find \( \lim _{x \rightarrow 0}(1000x \ln x) \). From the context, as \( x \to 0^+ \), the immediate observation is that \( x \to 0 \) and \( \ln x \to -\infty \). Together, they form a \( 0 \cdot -\infty \) scenario, making it a classic candidate for using l'Hôpital's Rule.Understanding the approaching behavior of functions is vital as it provides insights into function continuity and asymptotic trends, which are essential for further topics in calculus and real analysis.

Logarithmic Properties

Logarithms are powerful tools that help simplify multiplication operations into additions, and exponentiation into multiplication, thanks to their properties. They are especially useful when being combined with calculus, such as in problems involving limits and differentiation.One frequently used logarithmic property is \( \ln(a^b) = b \ln a \), which allows us to bring down exponents as coefficients. This feature was employed in the solution to transform the original term \( x \ln x^{1000} \) into \( 1000x \ln x \). This step was crucial for simplifying the problem and preparing it for the application of l'Hôpital's Rule.Understanding these properties allows us to manipulate expressions effectively, making them more manageable for computation and further analysis. Thus, whenever you encounter logarithms, recalling their main properties can greatly assist in analyzing complex expressions.