Problem 365 - Calculus Volume 1

Question

For the following exercises, determine whether you can apply L'Hôpital's rule directly. Explain why or why not. Then, indicate if there is some way you can alter the limit so you can apply L'Hópital's rule. $$ \lim _{x \rightarrow 0} \frac{x^{2}}{1 / x} $$

Step-by-Step Solution

Answer

No, L'Hôpital's rule cannot be applied directly since it's not an indeterminate form.

Step 1: Analyze the original limit

Examine the given limit \( \lim _{x \rightarrow 0} \frac{x^{2}}{1 / x} \). Substituting \( x = 0 \) gives an expression of \( \frac{0^2}{1/0} = \frac{0}{\infty} \). This is not an indeterminate form that L'Hôpital's rule can be directly applied to.

Step 2: Simplify the expression

Simplify the expression inside the limit: \( \frac{x^2}{1/x} = x^2 \cdot x = x^3 \). Now the limit becomes \( \lim_{x \rightarrow 0} x^3 \).

Step 3: Evaluate the simplified limit

The limit \( \lim_{x \rightarrow 0} x^3 = 0^3 = 0 \). Since the limit simplifies to a polynomial evaluation, L'Hôpital's rule is not applicable nor required here.

Key Concepts

Indeterminate Forms

Indeterminate forms appear in calculus when you try to evaluate a limit, but get an expression that doesn’t directly give you a clear answer, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms are problematic because they don't fit into standard arithmetic rules, and hence, special techniques must be used to resolve them. L'Hôpital's Rule is a popular method for tackling these troublesome forms. It states that if the limits of the numerator and denominator both approach 0 or ±∞, the limit of their ratio can be found by taking derivatives. However, not all cases involving infinity or zero are indeterminate. In the exercise given, substituting \( x = 0 \) yields \( \frac{0}{\infty} \), which is a determinate form where direct analysis reveals the limit does not exhibit the complexity suitable for L'Hôpital's Rule. This provides an important practice in identifying when these rules might initially appear necessary but aren’t needed—simplifying expression might provide a straightforward resolution instead.

Limits in Calculus

Limits are foundational in calculus and express the value that a function approaches as the input approaches some value. They help in understanding behaviors of functions at particular points or as variables tend to infinity. Calculus itself revolves deeply around the concept of limits, underpinning much of differential and integral calculus. In the exercise scenario, evaluating \( \lim_{x \rightarrow 0} \frac{x^{2}}{1/x} \) initially seems intimidating because of the potential volatility at \( x = 0 \). However, by simplifying the expression to \( x^3 \), you find the limit directly becomes 0, offering clarity and simplification through limit evaluation. Here is the process:

• First, check if the function behaves irregularly, such as approaching infinity or zero in problematic ways.
• If simplification reveals a non-problematic form, such as the polynomial \( x^3 \), then evaluate the limit traditionally.

Understanding when and how to apply limits effectively is crucial in addressing complex function behaviors.

Polynomial Limits

Polynomial limits involve polynomials, which are algebraic expressions made up of terms consisting of variables raised to whole-number powers and constant coefficients. Evaluating the limit of a polynomial generally becomes straightforward, as polynomials are continuous everywhere, and the limit at a point can be found by direct substitution. In the solution, reducing the fraction \( \frac{x^{2}}{1/x} \) to \( x^3 \) showcased a polynomial, which simplifies the task to direct evaluation by substitution. For \( x^3 \), as \( x \) approaches 0, the expression becomes \( 0^3 \), hence the limit is 0. When you simplify expressions to form polynomial limits, keep these guidelines in mind:

• Look for ways to rewrite complex expressions as more manageable polynomials.
• Use substitution directly to evaluate limits of polynomials.
• Check for continuity, as polynomials are smooth and differentiable throughout their domain.

With such clarity in simplifying polynomial limits, many seemingly tricky problems become straightforward, eliminating unnecessary complexities.