Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine whether the function is continuous at the given point \(c .\) If the function is not continuous, determine whether the discontinuity is removable or non-removable. $$ f(x)=\frac{\cos x}{x} ; c=0 $$

Step-by-Step Solution

Answer

The function is not continuous at \(c=0\) due to a removable discontinuity.

Step 1: Definition of Continuity

For a function \(f(x)\) to be continuous at a point \(c\), three conditions must be met: 1) \(f(c)\) is defined, 2) the limit \(\lim\limits_{x \to c} f(x)\) exists, 3) \(\lim\limits_{x \to c} f(x) = f(c)\). First, check if \(f(0)\) is defined.

Step 2: Evaluate f(c)

Substituting \(c = 0\) into \(f(x)\), we get \(f(0) = \frac{\cos 0}{0}\). As division by zero is undefined, \(f(0)\) is not defined. This suggests \(f(x)\) may be discontinuous at \(c = 0\).

Step 3: Find the Limit of f(x) as x approaches c

Calculate \(\lim\limits_{x \to 0} \frac{\cos x}{x}\). Since direct substitution leads to indeterminate form \(\frac{0}{0}\), use L'Hôpital's Rule. Differentiate the numerator and denominator: numerator \(\cos x\) differentiates to \(-\sin x\), and denominator \(x\) to \(1\). Thus, limit becomes \(\lim\limits_{x \to 0} \frac{-\sin x}{1} = 0\).

Step 4: Determine Type of Discontinuity

Since \(f(0)\) is undefined, but \(\lim\limits_{x \to 0} f(x) = 0\) exists, we have a hole at \(x = 0\). This is a removable discontinuity, indicating we can redefine \(f(0) = 0\) to make \(f(x)\) continuous at that point.

Key Concepts

Removable Discontinuity

When examining the continuity of a function, we may encounter points where the function seems to "break." This break is known as a discontinuity. Discontinuities can be classified as removable or non-removable, depending on the ability to "fix" them.

With removable discontinuities, the issue stems from a point where the function is not defined, but a limit exists. Essentially, these are points where we can "fill in" the gap to make the function continuous.

In the case of the function \(f(x) = \frac{\cos x}{x}\) at \(c = 0\), we found that \(f(0)\) is undefined, yet the limit \(\lim\limits_{x \to 0} \frac{\cos x}{x} = 0\) exists. Hence, the discontinuity is removable.

To "remove" the discontinuity, we can define \(f(0) = 0\), thus making the function continuous. This type of corrective measure is possible only because the limit exists at that point.

Limits and L'Hôpital's Rule

Understanding the behavior of functions as they approach certain points is crucial, and limits help us with that. Limits can sometimes be hard to calculate directly, especially when they involve indeterminate forms like \(\frac{0}{0}\). This is where L'Hôpital's Rule comes in handy.

L'Hôpital's Rule is specifically designed to handle indeterminate forms that occur when directly substituting into the limit results in expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It states that for these indeterminate cases, we can differentiate the numerator and denominator separately and then re-evaluate the limit:

• Differentiate the numerator.
• Differentiate the denominator.
• Compute the limit again with these new expressions.

In this exercise, the limit \(\lim\limits_{x \to 0} \frac{\cos x}{x}\) turned into \(\frac{0}{0}\). Applying L'Hôpital's Rule allowed us to reframe it as \(\lim\limits_{x \to 0} \frac{-\sin x}{1} = 0\), a much simpler form.

Indeterminate Forms

Indeterminate forms arise in calculus when limits give inconclusive or undefined outcomes. They indicate that more analysis using specific mathematical tools is needed to resolve the limit.

The most common indeterminate forms include:

• \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\), which can be tackled using L'Hôpital's Rule.
• Forms like \(0 \times \infty\), \(\infty - \infty\), and others that might need algebraic manipulation or advanced calculus techniques.

In our specific problem, substituting \(x = 0\) in \(\frac{\cos x}{x}\) yielded the \(\frac{0}{0}\) form. This initial form appears undefined, but with careful manipulation—via L'Hôpital's Rule or sometimes algebraic simplification—we can often find a definitive result.

Recognizing indeterminate forms is the starting point for choosing the right method to solve limits, which is crucial for analyzing function behavior at specific points.