Problem 7 - Analysis 1: Ein Lehr- und Arbeitsbuch für Studienanfänger

Question

Begründen sie, das die Funktionen \(f, g: R \rightarrow R\) mit \(f(x):= \begin{cases}\frac{\sin x}{x} & \text { für } x \neq 0 \\ 1 & \text { fïr } x=0\end{cases}\) und $$ g(x):= \begin{cases}\frac{\cos x-1}{x} & \text { für } x \neq 0 \\ 0 & \text { fis } x=0\end{cases} $$ stetig sind. Fast alle bisher in Beispielen und Aufgaben un= tersuchten Funktionen haben eine gemeinsame Eigenschaft: Sie sind in jedem Punkt ihres Definitionsbereiches differenzierbar. Solche Punktionen nennen wir differenzierbar.

Step-by-Step Solution

Answer

The functions \( f(x) \) and \( g(x) \) are continuous across their domains, including at \( x = 0 \).

Step 1: Definition and Continuity of Function f(x)

The function \( f(x) \) is defined as \( f(x) = \frac{\sin x}{x} \) for \( x eq 0 \) and \( f(0) = 1 \). For continuity at \( x = 0 \), we need \( \lim_{x \to 0} f(x) = f(0) = 1 \). Since we know the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), \( f(x) \) is continuous at \( x = 0 \). For other \( x eq 0 \), \( f(x) = \frac{\sin x}{x} \) is continuous as both \( \sin x \) and \( x \) are continuous, and division by non-zero \( x \) is allowed.

Step 2: Verification of Limit for f(x)

To reinforce that \( f(x) \) is continuous at \( x = 0 \), apply the limit: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). As both functions \( \sin x \) and \( x \) are differentiable around zero, \( f(x) \) satisfies the constraint \( \lim_{x \to 0} f(x) = f(0) \). Therefore, \( f(x) \) is continuous at \( x = 0 \).

Step 3: Definition and Continuity of Function g(x)

The function \( g(x) \) is defined as \( g(x) = \frac{\cos x - 1}{x} \) for \( x eq 0 \) and \( g(0) = 0 \). For \( g(x) \) to be continuous at \( x = 0 \), we need \( \lim_{x \to 0} g(x) = 0 \). Utilizing L'Hôpital's Rule, applied as the form \( \frac{0}{0} \) occurs, we take derivatives to get: \( \lim_{x \to 0} \frac{-\sin x}{1} = 0 \). This confirms \( g(x) \) is continuous at \( x = 0 \).

Step 4: Verification of Limit for g(x)

Reconfirm continuity by taking the limit \( \lim_{x \to 0} \frac{\cos x - 1}{x} = \lim_{x \to 0} \frac{-\sin x}{1} = 0 \). Therefore, \( \lim_{x \to 0} g(x) = g(0) = 0 \). Additionally, \( g(x) = \frac{\cos x - 1}{x} \) remains continuous for all \( x eq 0 \) as cosine is continuous and differentiable, confirming \( g(x) \) is fully continuous over its domain.

Key Concepts

Limits and Continuity

Understanding limits and continuity is essential for grasping how functions behave around a specific point. In mathematics, a function is said to be continuous at a point if the limit of the function as it approaches the point equals the function's value at that point.
For the function \( f(x) = \frac{\sin x}{x} \), which is defined as 1 when \( x = 0 \) and as \( \frac{\sin x}{x} \) for \( x eq 0 \), we need to determine its continuity at \( x = 0 \). We know from fundamental limits that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Since \( f(0) = 1 \), the function is continuous at \( x = 0 \).
Similarly, for the function \( g(x) = \frac{\cos x - 1}{x} \) defined as 0 when \( x = 0 \), continuity requires that \( \lim_{x \to 0} g(x) = 0 \). Using L'Hôpital's Rule, we can confirm that \( \lim_{x \to 0} \frac{-\sin x}{1} = 0 \), thus \( g(x) \) is also continuous at \( x = 0 \).
Both \( f(x) \) and \( g(x) \) demonstrate continuity over their domains because they are built from continuous trigonometric functions and maintain this continuity across the entire real line except at their defining points.

Differentiability

Differentiability is closely linked to continuity but is a stronger condition. For a function to be differentiable at a point, it must be continuous there. However, not all continuous functions are differentiable.
In simpler terms, a function is differentiable at a point if it has a defined tangent there. Looking at \( f(x) \), which is expressed as \( \frac{\sin x}{x} \) away from zero, shows differentiability because both \( \sin x \) and \( x \) are differentiable. At \( x = 0 \), the value of \( f(x) \) is set to match the limit, ensuring it remains flat, a necessary condition for differentiability.
For \( g(x) = \frac{\cos x - 1}{x} \), differentiability involves ensuring that as \( x \to 0 \), the function smoothly approaches zero. Using L'Hôpital’s Rule guarantees that the derivative exists at \( x = 0 \) because both the numerator and denominator change at compatible rates as \( x \to 0 \).
Despite being expressed in piecewise form, the behavior of both \( f(x) \) and \( g(x) \) makes them differentiable across their domains, further solidifying the robust nature of trigonometric functions in calculus.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool for evaluating limits that involve indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule states that if you have a limit approaching one of these forms, you can differentiate the top and bottom of the fraction separately and then re-evaluate the limit.
In the case of \( g(x) = \frac{\cos x - 1}{x} \), evaluating the limit directly as \( x \to 0 \) results in \( \frac{0}{0} \), an indeterminate form. By applying L'Hôpital's Rule, taking the derivative of the numerator \( -\sin x \) and the denominator \( 1 \), you get \( \lim_{x \to 0} \frac{-\sin x}{1} = 0 \). This process simplifies the finding and clarifies the function’s behavior near \( x = 0 \).
Always ensure that the prerequisites for using L'Hôpital's Rule—both functions being differentiable near the point in question—are met. This technique is incredibly useful not just in proving continuity but also in other challenging limit evaluations where traditional methods stall.