Problem 83 - 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. $$ \lim _{x \rightarrow \infty}\left(\frac{4 x^{2}+x-3}{2 x^{2}+1}\right) $$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Determine the Form of the Limit

First, substitute \( x = \infty \) into the expression \( \frac{4x^2 + x - 3}{2x^2 + 1} \). Notice that as \( x \rightarrow \infty \), both the numerator and the denominator approach infinity, leading to the indeterminate form \( \frac{\infty}{\infty} \).

Step 2: Differentiate the Numerator and Denominator

Using l'Hôpital's Rule, we differentiate the numerator \( f(x) = 4x^2 + x - 3 \) and the denominator \( g(x) = 2x^2 + 1 \). The derivative of the numerator \( f'(x) = 8x + 1 \) and the derivative of the denominator \( g'(x) = 4x \).

Step 3: Apply l'Hôpital's Rule

Substituting the derivatives back into the limit gives us \( \lim_{x \rightarrow \infty} \left( \frac{8x + 1}{4x} \right) \).

Step 4: Simplify the Expression

Factor \( x \) out of both terms in the numerator and the denominator to simplify: \( \frac{8x + 1}{4x} = \frac{8 + \frac{1}{x}}{4} \).

Step 5: Evaluate the Simplified Limit

Evaluate the limit \( \lim_{x \rightarrow \infty} \left( \frac{8 + \frac{1}{x}}{4} \right) = \frac{8 + 0}{4} = \frac{8}{4} = 2 \).

Key Concepts

Indeterminate Forms

When dealing with limits, you might encounter expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These are commonly referred to as indeterminate forms. The term "indeterminate" suggests that, based on the expression alone, the limit could lead to different possible outcomes. This uncertainty is due to the fact that dividing by zero or infinity can produce a range of values depending on the surrounding context of the functions.To resolve these problems, specifically indeterminate forms like \(\frac{0}{0}\) or \(\frac{\pm \infty}{\pm \infty}\), mathematical tools such as l'Hôpital's Rule become very helpful. This rule helps transform the complex form into something more manageable and computable by using the derivatives of the functions involved.
In using this rule, you differentiate the numerator and the denominator separately and then re-evaluate the limit. This process often leads to finding a limit that was initially uncertain.

Differentiable Functions

Differentiable functions are fundamental when applying l'Hôpital's Rule. A function is said to be differentiable at a point or over an interval if it has a derivative at that point, meaning the function's rate of change can be measured.
This rule leverages the concept of derivatives to solve limits that result in indeterminate forms. It allows us to replace the original functions with their derivatives, taking advantage of the more manageable expressions.For example, if you have functions \(f(x)\) and \(g(x)\) that are both differentiable, then when their ratio approaches an indeterminate form, you can apply l'Hôpital's Rule.
By differentiating \(f(x)\) to give \(f'(x)\) and \(g(x)\) to provide \(g'(x)\), we transform the original expression \(\frac{f(x)}{g(x)}\) into \(\frac{f'(x)}{g'(x)}\). This approach hinges on differentiable functions providing derivatives that are simpler to evaluate or resolve.

Limit Evaluation

Evaluating limits is a key mathematical process that helps us understand the behavior of functions as they approach specific points or infinity. In the context of l'Hôpital's Rule, it involves calculating the limit of a function ratio after transforming an indeterminate form using derivatives.Start by substituting the limit point into the expression. If this substitution results in an indeterminate form, apply l'Hôpital's Rule by differentiating the numerator and denominator.
Once differentiated, substitute back into the limit expression. Frequently, this leads to a simpler form that can be directly evaluated.
For instance, if the initial substitution in \(\frac{4x^2 + x - 3}{2x^2 + 1}\) results in \(\frac{\infty}{\infty}\), differentiating gives us \(\frac{8x + 1}{4x}\). Evaluating this simplified expression as \(x\) approaches infinity, leads to \(\frac{8 + 0}{4} = 2\). This systematic approach makes difficult problems tractable, revealing the true behavior of functions at challenging points.