Problem 18 - Single Variable Calculus

Question

evaluate the limit using l'Hôpital's Rule if appropriate. $$ \lim _{x \rightarrow 1} \frac{x^{1 / 2}-x^{1 / 3}}{x-1} $$

Step-by-Step Solution

Answer

The limit of the given function as x approaches 1 is \(\frac{1}{6}\) using l'Hôpital's Rule.

Step 1: Identify indeterminate form

First, let's determine if the function is indeterminate at x = 1: $$ \frac{1^{1 / 2}-1^{1 / 3}}{1-1} = \frac{1-1}{0} = \frac{0}{0} $$ The function takes the form of \(\frac{0}{0}\) when x approaches 1, which is an indeterminate form. So, we can proceed with l'Hôpital's Rule.

Step 2: Applying l'Hôpital's Rule

Apply l'Hôpital's Rule by taking the derivative of the numerator and denominator with respect to x: $$ \frac{d}{dx}(x^{1 / 2}-x^{1 / 3}) = \frac{1}{2}x^{-1 / 2} - \frac{1}{3}x^{-2 / 3} $$ $$ \frac{d}{dx}(x-1) = 1 $$ The new limit becomes: $$ \lim_{x \rightarrow 1} \frac{\frac{1}{2}x^{-1 / 2} - \frac{1}{3}x^{-2 / 3}}{1} $$

Step 3: Evaluating the Limit

Evaluating the new limit as x approaches 1: $$ \frac{\frac{1}{2}(1)^{-1 / 2} - \frac{1}{3}(1)^{-2 / 3}}{1} = \frac{1}{2} - \frac{1}{3} $$

Step 4: Simplifying the Answer

Now, simplify the answer: $$ \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} $$ So, the limit of the given function as x approaches 1 is \(\frac{1}{6}\).

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms are expressions that lack a clear or predictable value at certain points, typically during limit evaluations. They pose a challenge when one attempts to evaluate limits where the initial substitution results do not yield an obvious numerical value. One common type is \(\frac{0}{0}\), which arises from functions that have a numerator and denominator both approaching zero as \(x\) approaches a certain value.

When faced with an indeterminate form like \(\frac{0}{0}\), it indicates potential for using mathematical techniques such as l'Hôpital's Rule to find a definitive limit. This rule provides a method to resolve such forms by differentiating the numerator and denominator until a clear value can be identified.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. When working with l'Hôpital's Rule, derivatives become crucial. They are used to simplify the problem into one that’s manageable when the original function yields an indeterminate form.

In the given problem, the derivatives of the expressions \(x^{1/2} - x^{1/3}\) and \(x - 1\) are calculated. The differentiation process involves applying power rules to each term:

• The derivative of \(x^{1/2}\) is \(\frac{1}{2} x^{-1/2}\).
• The derivative of \(x^{1/3}\) is \(\frac{1}{3} x^{-2/3}\).
• The derivative of \(x - 1\) is simply 1, as linear terms differentiate to their coefficients.

These calculations transform the expression into a form that is easier to analyze when applying l'Hôpital's Rule.

Limit Evaluation

Limit evaluation is a central technique in calculus used to find the value that a function approaches as the input approaches a certain point. Using l'Hôpital's Rule helps simplify situations involving indeterminate forms.

Once you have derived the necessary derivatives, you can substitute back and evaluate the limit of these new expressions. In the exercise, after applying the derivatives, the limit transforms to:\[\lim_{x \rightarrow 1} \frac{\frac{1}{2}x^{-1 / 2} - \frac{1}{3}x^{-2 / 3}}{1}\]Substituting \(x = 1\) directly into this expression, it simplifies to \(\frac{1}{2} - \frac{1}{3}\), demonstrating a straightforward means of finding the limit once the function has been reorganized.

Calculus Problems

Calculus problems often involve complex computations, such as evaluating difficult limits. Challenges include handling functions that manifest as indeterminate forms, which require special strategies like l'Hôpital's Rule.

With knowledge of derivatives and limits, students can tackle calculus problems by:

• Identifying indeterminate forms that require transformation for evaluation.
• Finding and applying derivatives effectively, often simplifying expressions before substituting numbers.
• Performing careful arithmetic to obtain accurate final results.

Comprehending these concepts enables a student to approach calculus problems with confidence, transforming seemingly complex problems into simpler, solvable ones.