Problem 79 - 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. $$ \lim _{x \rightarrow-3}\left(\frac{x^{2}-9}{x+3}\right) $$

Step-by-Step Solution

Answer

The limit is \(-6\).

Step 1: Check for Indeterminate Form

Evaluate the limit by substituting \( x = -3 \) into the function \( \frac{x^2 - 9}{x+3} \). We have:\[\lim_{x \rightarrow -3}\left(\frac{x^2 - 9}{x + 3}\right) = \frac{(-3)^2 - 9}{-3 + 3} = \frac{9 - 9}{0} = \frac{0}{0}\]This is an indeterminate form, so we can apply l'Hôpital's Rule.

Step 2: Differentiate Numerator and Denominator

Apply l'Hôpital's Rule by differentiating the numerator and the denominator separately. The derivative of the numerator \( x^2 - 9 \) is \( 2x \), and the derivative of the denominator \( x + 3 \) is \( 1 \).

Step 3: Reevaluate the Limit

Substitute back into the limit with these derivatives:\[\lim_{x \rightarrow -3}\left(\frac{2x}{1}\right) = 2(-3) = -6\]

Step 4: Conclude with the Limit Value

The limit has been successfully calculated as \(-6\) after applying l'Hôpital's Rule.

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms are expressions that do not have a well-defined value or limit when evaluated directly. They often appear as a result of standard limit calculations. The most common indeterminate forms include \(\frac{0}{0}\) and \(\frac{\pm \infty}{\pm \infty}\). These forms present a challenge because direct substitution in a limit calculation leads to ambiguous outcomes. Fortunately, we have tools like l'Hôpital's Rule to help resolve these situations.

l'Hôpital's Rule provides a way to calculate the limit of functions that result in indeterminate forms by focusing on the derivatives of the numerator and the denominator. By transforming the original complex expression into a simpler form using differentiation, the limit can be evaluated more straightforwardly. For instance, if you encounter \(\lim_{x \rightarrow a} \left( \frac{f(x)}{g(x)} \right) = \frac{0}{0}\), l'Hôpital's Rule can be applied, provided the derivatives exist, to logically deduce the limit. This approach turns a seemingly impossible calculation into a manageable one.

Differentiation

Differentiation is a core concept of calculus that speaks to the concept of change. It is the process of finding the derivative of a function, which represents the rate at which a function is changing at any given point. When applying l'Hôpital's Rule, differentiation is the key technique used to resolve indeterminate forms.

In our example, when evaluating \( \lim_{x \rightarrow -3} \left( \frac{x^2 - 9}{x+3} \right) \), direct substitution leads to the indeterminate form \(\frac{0}{0}\). By differentiating the numerator and the denominator, we obtain their respective derivatives: \(2x\) for \(x^2 - 9\) and \(1\) for \(x + 3\). This transforms the problem into a new limit problem: \( \lim_{x \rightarrow -3} \left( \frac{2x}{1} \right) \). With this differentiation approach, we can now easily calculate the new limit and find that the solution is \(-6\).

This process highlights the power and significance of differentiation, as it enables us to simplify complex calculations and provides insight into how functions behave locally.

Limits in Calculus

The concept of limits is foundational in calculus because it lays the groundwork for understanding how functions behave as they approach a particular point. Limits allow us to handle cases where functions are undefined at certain points or where they behave unpredictably.

In the context of l'Hôpital's Rule and indeterminate forms, limits help us make sense of situations where both the numerator and the denominator approach zero or infinity. They guide us to calculate the behavior of functions even when direct evaluation is impossible.

Consider the limit \( \lim_{x \rightarrow -3} \left( \frac{x^2 - 9}{x+3} \right) \). At \(x = -3\), the direct computation results in \(\frac{0}{0}\), which is undefined. By understanding limits, we use the rule to transform this expression into a solvable form using derivatives. Concluding with \( \lim_{x \rightarrow -3} \left( \frac{2x}{1} \right) = -6\) showcases the effectiveness of limits in determining function behavior at challenging points.

Limits are not just theoretical tools; they provide practical solutions to real-world problems in mathematics, physics, engineering, and beyond, making them an essential skill in the toolkit of anyone studying calculus.