Problem 81 - Calculus and its applications

Question

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'Hôpital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 2}\left(\frac{x^{3}+5 x-18}{2 x^{2}-8}\right) $$

Step-by-Step Solution

Answer

The limit is \( \frac{17}{8} \).

Step 1: Check for Indeterminate Form

First, substitute \(x = 2\) into the function to determine if it is an indeterminate form. We have \( f(x) = x^3 + 5x - 18 \) and \( g(x) = 2x^2 - 8 \). Calculate \( f(2) = 2^3 + 5 \cdot 2 - 18 = 8 + 10 - 18 = 0 \). Similarly, calculate \( g(2) = 2 \cdot 2^2 - 8 = 8 - 8 = 0 \). At \( x = 2 \), \( \frac{f(x)}{g(x)} = \frac{0}{0} \), which is an indeterminate form.

Step 2: Apply L'Hôpital's Rule

Because the limit forms an indeterminate \( \frac{0}{0} \), we can apply L'Hôpital's Rule by differentiating the numerator and the denominator. Differentiate the numerator: \( f'(x) = \frac{d}{dx}(x^3 + 5x - 18) = 3x^2 + 5 \). Differentiate the denominator: \( g'(x) = \frac{d}{dx}(2x^2 - 8) = 4x \).

Step 3: Rewrite the Limit Using Derivatives

Replace the original function with its derivatives as per L'Hôpital's Rule. Therefore, the limit becomes: \[ \lim_{x \to 2} \frac{3x^2 + 5}{4x} \].

Step 4: Evaluate the New Limit

Substitute \(x = 2\) in the derivative-based limit: \( \frac{3(2)^2 + 5}{4 \cdot 2} = \frac{3 \cdot 4 + 5}{8} = \frac{12 + 5}{8} = \frac{17}{8} \).

Step 5: Conclusion: State the Limit

Therefore, the limit \( \lim_{x \to 2} \frac{x^3 + 5x - 18}{2x^2 - 8} = \frac{17}{8} \).

Key Concepts

Understanding Indeterminate Forms

Indeterminate forms arise in calculus when substituting a particular value into a function results in undefined or ambiguous expressions, such as \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \). These forms signal that a direct evaluation of the limit is inconclusive, which can leave many students puzzled. Rather than giving up, we use techniques like L'Hôpital's Rule to simplify the expression.

It's crucial to first check for these indeterminate forms by substituting the value at which you're evaluating the limit. For example, if you insert \( x = 2 \) into both the numerator and denominator of our function \( \frac{f(x)}{g(x)} = \frac{x^3 + 5x - 18}{2x^2 - 8} \), and both evaluate to zero, you'll get \( \frac{0}{0} \). This confirms that the expression is indeterminate, signaling the need for another strategy for evaluation.

Basics of Differentiation

Differentiation, one of the core operations in calculus, involves finding the derivative or the instantaneous rate of change of a function. This process transforms a function into its derivative by applying certain rules and principles, like the power rule, product rule, or chain rule.

When dealing with limits, especially those in the indeterminate form, we often use differentiation to simplify the original expression for easier computation. In our exercise, after establishing the indeterminate form, both the numerator \( f(x) = x^3 + 5x - 18 \) and the denominator \( g(x) = 2x^2 - 8 \) were differentiated. Applying the respective derivative formulas, we found \( f'(x) = 3x^2 + 5 \) and \( g'(x) = 4x \). This step is pivotal in transforming the original problem into a more manageable expression for limit evaluation.

Insights into Limits

The concept of limits is foundational in understanding calculus and involves determining the value that a function approaches as the input approaches a certain point. When faced with indeterminate forms, limits can be particularly tricky yet rewarding to dissect.

In our example, we needed to find \( \lim_{x \to 2} \frac{x^3 + 5x - 18}{2x^2 - 8} \), but since directly evaluating at \( x=2 \) results in \( \frac{0}{0} \), the limit is unclear. This is where using L'Hôpital's Rule comes into play. We replace the original fractions with their derivatives and then evaluate the limit again, transforming a complex problem into something tangible that can now be computed as \( \frac{17}{8} \).

This process highlights how limits can reveal behavior of functions around points where direct substitution is impossible.

Solving Calculus Problems with L'Hôpital's Rule

L'Hôpital's Rule is a powerful method in calculus for dealing with problematic indeterminate forms. By using differentiation, it turns tough limits into solvable ones.

When a limit is in an indeterminate form, L'Hôpital's Rule provides a structured way to re-evaluate by taking the derivatives of both the numerator and the denominator. Once this is done, the limit problem often becomes straightforward. The rule is especially helpful in situations where other techniques or algebraic manipulation would fall short.

For our exercise, applying L'Hôpital's Rule turned a \( \frac{0}{0} \) scenario into a simple derivative evaluation: \( \lim_{x \to 2} \frac{3x^2 + 5}{4x} \). Upon evaluating this transformed limit, we obtain the final result of \( \frac{17}{8} \). Thus, L'Hôpital's Rule not only simplifies the process but also makes tricky calculus problems approachable.