Problem 121 - Calculus of a Single Variable

Question

Making a Function Continuous Let $$ f(x)=\frac{\sqrt{x+c^{2}}-c}{x}, \quad c > 0 $$

Step-by-Step Solution

Answer

The function \(f(x)=\frac{\sqrt{x+c^{2}}-c}{x}\) is continuous at x=0 and the limit as x approaches 0 is \(\frac{1}{2c}\).

Step 1: Identify the Given Function

We have been given a function, \(f(x)=\frac{\sqrt{x+c^{2}}-c}{x}\). This is a rational function. The goal here is to make this function continuous. For a rational function to be continuous, the denominator should not be equal to zero.

Step 2: Evaluate the Limit as X Approaches 0

To find out if our given function is continuous, we will have to evaluate the limit of the function as x approaches 0, i.e., \(\lim_{{x\to 0}} \frac{\sqrt{x+c^{2}}-c}{x}\). If this limit exists and equals \(f(0)\), then the function is continuous at x=0.

Step 3: Simplify the Rational Function

The numerator of our function in the limit looks like a difference of squares which can be simplified by rationalizing. Let's multiply the numerator and denominator by the conjugate of the numerator, i.e., \(\sqrt{x+c^{2}}+c\). Thus, the expression becomes, \(\lim_{{x\to 0}} \frac{\sqrt{x+c^{2}}-c}{x} \cdot \frac{\sqrt{x+c^{2}}+c}{\sqrt{x+c^{2}}+c}\). Simplifying this, we get \(\lim_{{x\to 0}} \frac{x}{x(\sqrt{x+c^{2}}+c)}\), which simplifies to \(\lim_{{x\to 0}} \frac{1}{\sqrt{x+c^{2}}+c}\).

Step 4: Evaluate the Limit

The expression now left is in a form that allows the substitution x=0 directly. Substituting x=0, we get \(\frac{1}{\sqrt{0+c^{2}}+c}\) which simplifies to \(\frac{1}{2c}\). This shows that the limit as x approaches 0 for our function is \(\frac{1}{2c}\). As the limit is defined, the function is continuous at x=0.

Key Concepts

Rational Function Continuity

Understanding the continuity of rational functions, such as the given function \(f(x)=\frac{\sqrt{x+c^{2}}-c}{x}\), where \(c > 0\), is critical in calculus. A rational function is a ratio of two functions where the denominator is not equal to zero; thus, such functions can be discontinuous if their denominators become zero at any point in their domain.

Continuity at a point means that the function is defined at that point, the limit as you approach the point exists, and that the limit equals the function's value at that point. For the function \(f(x)\) to be continuous at \(x = 0\), we need to ensure the function is defined and the limit as \(x\) approaches 0 exists and equals \(f(0)\).

Limits in Calculus

Limits are a foundational concept in calculus, which deal with the behavior of functions as they approach a certain point. The expression \(\lim_{{x\to a}} f(x)\) represents the limit of the function \(f(x)\) as \(x\) approaches \(a\). If this limit is equal to the function's value at \(a\), \(f(a)\), the function is continuous at \(a\).

To determine the continuity of \(f(x)\), we evaluate \(\lim_{{x\to 0}} \frac{\sqrt{x+c^{2}}-c}{x}\). The existence of this limit is crucial to establish whether or not there is a point of discontinuity at \(x = 0\).

Rationalizing Numerators

Rationalizing the numerator or denominator of a fraction is a technique used to eliminate radicals by multiplying by an appropriate conjugate. In our example, multiplying the numerator and denominator of the function by the conjugate of the numerator allows us to utilize the difference of squares identity, turning a complex radical expression into a simpler one.

When we multiply by the conjugate \(\sqrt{x+c^{2}}+c\), the numerator becomes a difference of squares, which simplifies to \(x\), and this simplification is what eventually allows us to evaluate the limit as \(x\) approaches 0. This step is crucial for determining the continuity of functions involving square roots and other radical terms.

Difference of Squares

The difference of squares is a mathematical shortcut used to simplify expressions. It comes from the algebraic identity \(a^2 - b^2 = (a + b)(a - b)\). In our exercise, after rationalizing the numerator, we end up with such a difference of squares: \((\sqrt{x+c^{2}})^2 - c^2\), which simplifies to \(x\).

This simplification is particularly useful when dealing with limits, as it often transforms an indeterminate form into one that can be directly evaluated. By understanding and applying the difference of squares identity, the problem of finding the limit of our function as \(x\) approaches 0 was made significantly more manageable.