Question

By use of L'Hôpital's rule, show that (a) \(\lim _{x \rightarrow \infty} \frac{x^{n}}{e^{x}}=0\) (b) \(\lim _{x \rightarrow 0^{+}} x \ln x=0\) (c) \(\lim _{x \rightarrow 0^{+}} x^{x}=1\)

Short answer

(a) 0, (b) 0, (c) 1

Step-by-step solution

Identifying L'Hôpital's Rule

L'Hôpital's Rule applies to limits that are in an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). We can use it to find \( \lim _{x \rightarrow \infty} \frac{x^{n}}{e^{x}} \) because both numerator and denominator approach infinity as \( x \to \infty \).

Applying L'Hôpital's Rule to (a)

Differentiate the numerator \( x^n \) and the denominator \( e^x \) once: the derivative of \( x^n \) is \( nx^{n-1} \) and the derivative of \( e^x \) is \( e^x \). The limit is then \( \lim_{x \to \infty} \frac{nx^{n-1}}{e^x} \). Repeat this process \( n \) times. Ultimately, the limit becomes \( \lim_{x \to \infty} \frac{n!}{e^x} = 0 \) because \( e^x \) grows faster than any polynomial.

Setting up (b) x ln x

Rewrite \( x \ln x \) as \( \frac{\ln x}{\frac{1}{x}} \). As \( x \to 0^+ \), \( \ln x \to -\infty \) and \( \frac{1}{x} \to \infty \), leading to an indeterminate form \( \frac{-\infty}{\infty} \).

Applying L'Hôpital's Rule to (b)

Differentiate the numerator \( \ln x \) and the denominator \( \frac{1}{x} \): the derivative of \( \ln x \) is \( \frac{1}{x} \) and the derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \). Thus, the limit becomes \( \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} -x = 0 \).

Simplifying (c) x^x using logarithms

Write \( x^x \) as \( e^{x \ln x} \). We need to find \( \lim_{x \to 0^+} x \ln x = 0 \) from part (b). Then, \( \lim_{x \to 0^+} x^x = e^0 = 1 \).

Conclusion and Verification

Verify each calculation to ensure correctness. L'Hôpital's Rule was correctly applied, and simplifying using logarithm properties gave valid results. Therefore, the solutions hold true.

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a crucial tool in calculus for handling limits that take on indeterminate forms. These forms include expressions like \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\), situations where direct evaluation of limits isn't straightforward. Here's how you can use it effectively:

• First, confirm the limit expression results in an indeterminate form.
• Take the derivative of both the numerator and the denominator of the fraction.
• Re-evaluate the limit using these new expressions.

For example, in the expression \(\lim _{x \rightarrow \infty} \frac{x^{n}}{e^{x}}\), both numerator and denominator approach infinity as \(x\) increases. Applying L'Hôpital's Rule involves differentiating \(x^n\) and \(e^x\) step-by-step until the expression is no longer indeterminate. Eventually, you'll find that \(e^x\) grows more rapidly than any polynomial, leading the limit to converge to zero.

Limits and Continuity

The concepts of limits and continuity form the backbone of calculus, providing the framework for analyzing the behavior of functions as inputs approach specific points.

• Limit: This tells you what value a function approaches as the input nears a particular point. It's essential for evaluating expressions that aren't readily computable through simple substitution.
• Continuity: A function is continuous at a point if the limit exists and equals the function's value at that point. This means you can draw the function without lifting your pencil at that spot.

For instance, consider \(\lim _{x \to 0^+} x \ln x\). As \(x\) approaches zero from the right, \(\ln x\) heads to negative infinity. But if you transform this into a form amenable for L'Hôpital's Rule, such as \(\frac{\ln x}{\frac{1}{x}}\), you can carefully find the limit and ensure continuity in a new form, allowing for a solution like zero that might not seem apparent initially.

Derivative Calculations

Derivative calculations serve as a fundamental tool in analyzing the rate of change of functions. When you need to use L'Hôpital's Rule, derivatives become your go-to technique.

• Power Rule: Use to differentiate functions like \(x^n\), where the result becomes \(nx^{n-1}\).
• Exponential Functions: The derivative of \(e^x\) remains \(e^x\), a neat property that simplifies many calculations.

When dealing with something like \(\lim _{x \to 0^+} x \ln x\) rewritten as \(\frac{\ln x}{\frac{1}{x}}\), you'd differentiate to find \(\frac{1}{x}\) for \(\ln x\) and \(-\frac{1}{x^2}\) for \(\frac{1}{x}\). Subsequently, using these derivatives helps to simplify the limit's calculation further and achieve a precise result.