Question

Determine the largest interval over which the given function is continuous. $$ g(x)=\left\\{\begin{array}{ll} \frac{\sin x}{x}, & x \neq 0 \\ 0, & x=0 \end{array}\right. $$

Short answer

The function is continuous on \( (-\infty, 0) \cup (0, \infty) \).

Step-by-step solution

Identify the Points of Discontinuity

The given function \( g(x) \) is defined piecewise. For \( x eq 0 \), the function is \( \frac{\sin x}{x} \) and at \( x = 0 \), the function is defined as 0 to make \( g(x) \) continuous at the origin. This change of definition might indicate potential discontinuity at \( x = 0 \).

Verify Continuity at \( x \neq 0 \)

For all \( x eq 0 \), \( \frac{\sin x}{x} \) is continuous, as it is composed of continuous functions sine and linear division where the divisor is non-zero. So, \( g(x) \) is continuous for all \( x eq 0 \).

Analyze Continuity at \( x = 0 \)

We must check if the function is continuous at \( x = 0 \). To do this, we compute the limit \( \lim_{x \to 0} \frac{\sin x}{x} \). Using L'Hopital's rule or the standard limit, we find that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). This does not equal the defined value of the function at \( x = 0 \), which is 0. Therefore, \( g(x) \) is not continuous at \( x = 0 \).

Determine the Largest Interval of Continuity

Since \( g(x) \) is not continuous at \( x = 0 \) but is continuous everywhere else as determined in Step 2, the largest interval over which \( g(x) \) is continuous is \( (-\infty, 0) \cup (0, \infty) \).

Key concepts

Piecewise Function

A piecewise function is a type of function that has different expressions based on different intervals of the domain. These functions are designed to address specific conditions within their domains. In the given exercise, our function \( g(x) \) is piecewise, consisting of two parts:

• For \( x eq 0 \), \( g(x) = \frac{\sin x}{x} \).
• At \( x = 0 \), \( g(x) = 0 \).

The reason for using a piecewise definition here is that \( \frac{\sin x}{x} \) is not defined at \( x = 0 \) since division by zero is undefined. Hence, a specific value (0 in this case) is assigned to the function at that point to try addressing continuity issues.

Limits

Limits are crucial for understanding the behavior of functions at points where they might not be explicitly defined. A limit examines where a function is trending as it approaches a particular point.
In our exercise, we calculate the limit \( \lim_{x \to 0} \frac{\sin x}{x} \) to assess the behavior of the function near 0. Using a standard trigonometric limit, we know that this particular limit equals 1.
Limits are especially helpful when considering continuity of piecewise functions because they help determine the intended value at points where the function's direct evaluation may result in indeterminacy.

Intervals of Continuity

An interval of continuity is a range of input values for which a function is continuous. A function is considered continuous over an interval if it has no breaks, jumps, or points of discontinuity within that range.
In the context of \( g(x) \), we determined that it is continuous for all \( x eq 0 \). Therefore, its intervals of continuity exclude the point \( x = 0 \), giving us \( (-\infty, 0) \cup (0, \infty) \). These intervals indicate that aside from the singular discontinuity at zero, the function smoothly operates for all other real numbers.

l'Hôpital's Rule

l'Hôpital's rule is a powerful tool for evaluating limits that result in indeterminate forms like \( \frac{0}{0} \). The rule states if the limits of the numerator and denominator of a fraction both approach zero or infinity, then the limit of the fraction can be found by taking the derivative of the numerator and the derivative of the denominator separately, then finding the limit of their quotient.
In the exercise at hand, we considered the limit \( \lim_{x \to 0} \frac{\sin x}{x} \). Although this specific limit can be found using a standard result, it can also be found via l'Hôpital's rule:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( x \) is 1.

Therefore, \( \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \cos(0) = 1 \). This confirms that the function approaches 1 as \( x \to 0 \), which highlights the discontinuity at \( x = 0 \) where the function is set to 0.