Problem 75 - 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. $$ \lim _{x \rightarrow 5}\left(\frac{x^{2}-25}{2 x-10}\right) $$

Step-by-Step Solution

Answer

The limit is 5.

Step 1: Initial Substitution

First, attempt to directly substitute the value of \(x = 5\) into the function \(\frac{x^2 - 25}{2x - 10}\) to see if it results in an indeterminate form. Applying this substitution, we have \(\frac{5^2 - 25}{2(5) - 10} = \frac{25 - 25}{10 - 10} = \frac{0}{0}\). The expression yields \(\frac{0}{0}\), which is an indeterminate form, indicating we can apply l'Hôpital's Rule.

Step 2: Differentiate Numerator and Denominator

Calculate the derivatives of the numerator \(x^2 - 25\) and the denominator \(2x - 10\). The derivative of \(x^2 - 25\) is \(2x\), and the derivative of \(2x - 10\) is 2. So, we have \(\frac{d}{dx}(x^2 - 25) = 2x\) and \(\frac{d}{dx}(2x - 10) = 2\).

Step 3: Apply l'Hôpital's Rule

By l'Hôpital's Rule, replace the initial limit with the limit of the derivatives. Thus, we need to evaluate \(\lim_{x \to 5} \left(\frac{2x}{2}\right)\). Simplify the expression to \(\lim_{x \to 5} (x)\).

Step 4: Evaluate the Simplified Limit

Substitute \(x = 5\) into the simplified expression \(x\). This gives \(\lim_{x \to 5} (x) = 5\). Thus, \(\lim_{x \to 5} \left(\frac{x^2 - 25}{2x - 10}\right) = 5\).

Key Concepts

Indeterminate Forms

In calculus, when we talk about limits, an indeterminate form is a mathematical expression that is not immediately possible to evaluate directly. Two common types of indeterminate forms are \( \frac{0}{0} \) and \( \frac{\pm \infty}{\pm \infty} \). These forms often appear when directly substituting a value into a function, especially when solving limits of rational functions. They are called 'indeterminate' because their values cannot be determined from the form alone.
For instance, when a function's limit results in the form \( \frac{0}{0} \), it suggests that both the numerator and the denominator of the fraction become zero simultaneously as \( x \) approaches a specific value. This situation implies ambiguity, because the zero in both parts can arise in different ways that influence how the expression behaves nearby.
Indeterminate forms signify that the limit may exist, but we need to do more work to find it. This is where techniques like l'Hôpital's Rule become valuable.

Differentiation

Differentiation is a fundamental concept in calculus involving the calculation of a derivative, which represents the rate of change of a function's output with respect to its input. In the context of limits, and especially l'Hôpital's Rule, differentiation allows us to transform an indeterminate form into a determinate one that can be more easily evaluated.
The process involves taking the derivative of the function's numerator and the derivative of its denominator separately. For instance, if we have a function \( \frac{f(x)}{g(x)} \) and it is yielding an indeterminate form at some point \( x = a \), finding the derivatives \( f'(x) \) and \( g'(x) \) helps accordingly. This transforms the original limit problem into one that might be simpler to solve.
In the solved exercise, the function \( \frac{x^2 - 25}{2x - 10} \) resulted in \( \frac{0}{0} \) upon substitution. Differentiating the numerator, \( x^2 - 25 \), gives \( 2x \). For the denominator, \( 2x - 10 \), the derivative is a constant 2. These derivatives then allow us to simplify and solve the limit.

Limit Evaluation

Limit evaluation is the process of finding the value that a function approaches as its input gets arbitrarily close to a certain point. It is a key concept in calculus, often involving algebraic manipulation and special rules like l'Hôpital's Rule in the presence of indeterminate forms.
Successful limit evaluation might start with direct substitution to check for simple results. If an indeterminate form is encountered, techniques such as factorization, conjugation, or, notably, l'Hôpital's Rule are employed. In this rule, if direct substitution into the original function leads to \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \), we take the derivatives of the numerator and denominator. We then substitute into this new, simplified expression and evaluate the limit.
For example, in the exercise, after differentiating the numerator and the denominator, we simplified the expression to \( \lim_{x \to 5} \left( \frac{2x}{2} \right) = \lim_{x \to 5} (x) \). Evaluating this straightforward expression yielded the limit \( 5 \), effectively solving our original problem by making it approachable and clear.