Problem 77 - 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. $$ \lim _{x \rightarrow 1}\left(\frac{x^{3}+2 x-3}{x^{2}-1}\right) $$

Step-by-Step Solution

Answer

The limit is \( \frac{5}{2} \).

Step 1: Check for Indeterminate Form

First, substitute \( x = 1 \) directly into the expression \( \frac{x^3 + 2x - 3}{x^2 - 1} \). Calculate:\( x^3 + 2x - 3 = 1^3 + 2(1) - 3 = 0 \). \( x^2 - 1 = 1^2 - 1 = 0 \).Since both the numerator and the denominator are 0, the expression is in the indeterminate form \( \frac{0}{0} \).

Step 2: Apply l'Hôpital's Rule

Since the limit is initially \( \frac{0}{0} \), we can apply l'Hôpital's Rule, which involves differentiating the numerator and the denominator separately.

Step 3: Differentiate the Numerator and Denominator

Differentiate the numerator: \( f(x) = x^3 + 2x - 3 \). So, \( f'(x) = 3x^2 + 2 \). Differentiate the denominator: \( g(x) = x^2 - 1 \). So, \( g'(x) = 2x \).

Step 4: Evaluate the Limit of the Differentiated Function

The limit to evaluate is now:\[ \lim_{x \rightarrow 1} \left(\frac{f'(x)}{g'(x)}\right) = \lim_{x \rightarrow 1} \left(\frac{3x^2 + 2}{2x}\right)\]Substitute \( x = 1 \) into the differentiated functions:\( \frac{3(1)^2 + 2}{2(1)} = \frac{3 + 2}{2} = \frac{5}{2} \).

Step 5: Conclude the Result

Therefore, the limit of the original expression is the same as the limit of the differentiated function:\[ \lim_{x \rightarrow 1}\frac{x^3 + 2x - 3}{x^2 - 1} = \frac{5}{2} \]

Key Concepts

Indeterminate Forms

When evaluating limits, sometimes initial substitution of the limit point into the given function doesn't provide a clear answer. You might encounter expressions such as \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \). These are known as indeterminate forms.

Indeterminate forms essentially mean that the expression doesn't directly lead to a valid mathematical result, purely from substitution. This happens because the values approach zero or infinity, but it doesn't tell us how they approach these values.

Various techniques are employed to solve limits involving indeterminate forms. L'Hôpital's Rule is one of the most systematic approaches. Before applying this rule, always verify that the function actually results in an indeterminate form after substitution. This step is critical, as not every limit is indeterminate.

Identifying these forms helps us know where and why advanced tools, such as l'Hôpital's Rule, need to be used to resolve a limit successfully.

Limit Evaluation

Evaluating limits is a fundamental concept in calculus used to understand the behavior of functions as they approach specific points. In particular, when we have an indeterminate form like \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty}\), we want to know what value the function approaches as \( x \) gets closer to a point.

To evaluate these types of limits, l'Hôpital's Rule is a powerful tool. The rule states that if you have an indeterminate form after direct substitution, you can differentiate the numerator and the denominator separately. After differentiating, you then re-evaluate the limit of the resulting fraction.

For example, if \( \lim_{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right) \) results in \( \frac{0}{0} \), you can find the limit \( \lim_{x \rightarrow a}\left(\frac{f'(x)}{g'(x)}\right) \), assuming this new limit exists. The key to successful limit evaluation using this rule is ensuring that each step is done carefully, checking for existence and employing differentiation properly.

Differentiation

Differentiation is a core concept in calculus that involves finding the rate at which a function is changing at any given point. When applying l'Hôpital's Rule, differentiation takes center stage.

Let's break down the differentiation process:

• Identify the numerator \( f(x) \) and denominator \( g(x) \) of your function.
• Find the derivative of the numerator, which is \( f'(x) \).
• Find the derivative of the denominator, which is \( g'(x) \).

By replacing the original function with these derivatives, you're essentially comparing how each part of the function changes as you approach the limit.

For instance, in the original exercise, you first differentiate \( f(x) = x^3 + 2x - 3 \) to get \( f'(x) = 3x^2 + 2 \), and \( g(x) = x^2 - 1 \) to get \( g'(x) = 2x \). These derivatives are then used to re-evaluate the limit. Differentiation is a precise tool that uncovers deeper insights into the function's behavior, offering clarity where direct substitution fails.