Problem 76 - 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. $$ \lim _{x \rightarrow-2}\left(\frac{x^{2}-4}{x+2}\right) $$

Step-by-Step Solution

Answer

The limit is -4.

Step 1: Check for Indeterminate Form

Substitute \(x = -2\) directly into the fraction \(\frac{x^2 - 4}{x + 2}\). This becomes \(\frac{(-2)^2 - 4}{-2 + 2} = \frac{4 - 4}{0} = \frac{0}{0}\), which is an indeterminate form.

Step 2: Differentiate the Numerator and Denominator

Find the derivative of the numerator \(x^2 - 4\) and the denominator \(x + 2\). The derivative of the numerator \(x^2 - 4\) is \(2x\). The derivative of the denominator \(x + 2\) is \(1\).

Step 3: Apply L'Hôpital's Rule

Using L'Hôpital's Rule, apply the derivatives from Step 2 to the limit expression: \(\lim_{x \rightarrow -2}\left(\frac{x^2 - 4}{x + 2}\right) = \lim_{x \rightarrow -2}\left(\frac{2x}{1}\right)\).

Step 4: Compute the Limit

Substitute \(x = -2\) into the new expression from Step 3: \(\lim_{x \rightarrow -2}\left(\frac{2x}{1}\right) = \frac{2(-2)}{1} = -4\).

Key Concepts

Indeterminate Forms

Indeterminate forms arise in limit problems where the direct substitution gives uncertain expressions like \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \). These forms do not provide a straightforward answer as they are essentially undefined. Instead of directly reaching a conclusion, we explore further to find a meaningful limit.
L'Hôpital's Rule is particularly useful in handling these situations. It gives a systematic method to evaluate the limit by differentiating both the numerator and the denominator until the indeterminate form is resolved. To use L'Hôpital's Rule effectively, it is crucial to identify the presence of an indeterminate form initially through substitution or analysis.
If after substitution the result is indeterminate, this is a perfect scenario to apply L'Hôpital's Rule to find the true limit.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative of a function at any given point measures how the function's value changes as the input changes. It is a fundamental concept in calculus that plays a critical role in L'Hôpital's Rule.
When you encounter an indeterminate form, taking derivatives of the numerator and the denominator is the next step if you apply L'Hôpital's Rule. For example, consider a function \( f(x) = x^2 - 4 \). Its derivative, using basic differentiation rules, is \( 2x \). Similarly, the derivative of a linear function like \( g(x) = x + 2 \) is simply \( 1 \), because the rate of change is constant across all values of \( x \).
These derivatives help simplify the original limit expression, which can be evaluated with ease, thereby making differentiation a powerful tool in solving limit problems like these.

Limit Calculation

Limit calculation is the process of finding the value a function approaches as the input approaches some value. Understanding limits is essential, particularly when dealing with continuous functions and when direct substitution doesn't work due to indeterminate forms.
When applying L'Hôpital's Rule, once you have the derivatives from differentiation, you re-evaluate the limit using these derivatives. Substitute the limit point into the new, simplified fraction from Step 2 of L'Hôpital’s Rule. For example, with the derivatives obtained (from \( \frac{x^2-4}{x+2} \) transformed to \( \frac{2x}{1} \)), substitute \( x = -2 \) to find \( \frac{2(-2)}{1} = -4 \).
By successfully applying these steps, we can resolve indeterminate forms and accurately compute limits, gaining insights into the behavior of functions at specific points.