Problem 52 - Thomas Calculus

Question

$$\quad \text { Let } f(x)=\left\\{\begin{array}{ll}{x^{2} \sin (1 / x),} & {x < 0} \\ {\sqrt{x},} & {x > 0}\end{array}\right.$$ Find (a) \(\lim _{x \rightarrow 0^{+}} f(x)\) and (b) \(\lim _{x \rightarrow \sigma} f(x) ;\) then use limit definitions to verify your findings. (c) Based on your conclusions in parts (a) and (b), can you say anything about lim \(_{x \rightarrow 0} f(x) ?\) Give reasons for your answer.

Step-by-Step Solution

Answer

Both one-sided limits yield zero, so \( \lim_{x \to 0} f(x) = 0 \).

Step 1: Understanding the Piecewise Function

The function \( f(x) \) is defined as a piecewise function: \( f(x) = x^2 \sin(1/x) \) for \( x < 0 \) and \( f(x) = \sqrt{x} \) for \( x > 0 \). We need to analyze each piece separately and consider their behavior as \( x \) approaches zero from different directions.

Step 1: Evaluate \( \lim_{x \to 0^{+}} f(x) \)

When \( x > 0 \), \( f(x) = \sqrt{x} \). We calculate \( \lim_{x \to 0^{+}} \sqrt{x} \). As \( x \to 0^{+} \), \( \sqrt{x} \to 0 \). Thus, \( \lim_{x \to 0^{+}} f(x) = 0 \).

Step 2: Analyze \( \lim_{x \to 0^{-}} f(x) \)

When \( x < 0 \), \( f(x) = x^2 \sin(1/x) \). Since \( \sin(1/x) \) is bounded between -1 and 1, \( |x^2 \sin(1/x)| \leq x^2 \). As \( x^2 \to 0 \) when \( x \to 0^{-} \), \( x^2 \sin(1/x) \to 0 \) as well. So, \( \lim_{x \to 0^{-}} f(x) = 0 \).

Step 3: Conclusion for \( \lim_{x \to 0} f(x) \)

Since \( \lim_{x \to 0^{+}} f(x) = 0 \) and \( \lim_{x \to 0^{-}} f(x) = 0 \), the two one-sided limits are equal. Therefore, \( \lim_{x \to 0} f(x) = 0 \). The function is continuous at \( x = 0 \) based on these limits.

Key Concepts

Limits and Their Importance

In calculus, a limit help us understand the behavior of a function as the input approaches a certain value. They are crucial in determining the existence of values that the function nears but might not exactly reach. For our piecewise function, we examined the limit as the variable, \(x\), neared zero from the positive side and the negative side.

• When \(x > 0\), our function is \(\sqrt{x}\). As \(x\) approaches zero from the right, the square root of a small positive number gets closer to zero. Hence, \(\lim_{x \to 0^{+}} f(x) = 0\).
• When \(x < 0\), our function is \(x^2 \sin(1/x)\). Here, \(\sin(1/x)\) oscillates between -1 and 1, creating a challenge for exact limit prediction. However, \(x^2\) diminishes quickly to zero, meaning the entire term \(x^2 \sin(1/x)\) also nears zero. Thus, \(\lim_{x \to 0^{-}} f(x) = 0\).

By calculating these limits from both directions, we establish the overall limit at zero. Hence, \(\lim_{x \to 0} f(x) = 0\), linking to the concept of function behavior at specific points.

Understanding Piecewise Functions

A piecewise function is defined by different expressions depending on the input value. This is a versatile way to express function behavior that changes across different intervals. In our exercise, the piecewise function is written as:

• \(f(x) = x^2 \sin(1/x)\) for \(x < 0\).
• \(f(x) = \sqrt{x}\) for \(x > 0\).

Each piece of this function is analyzed separately to understand its general trend and behavior. With \(x^2 \sin(1/x)\), the function deals with oscillations due to the sine component, but the colder \(x^2\) coefficient plays a role in pushing it towards zero for small values of \(x\) leftwards. Similarly, \(\sqrt{x}\) provides direct insight as it smoothly approaches zero from positive inputs.
Piecewise functions require careful observation of each segment independently, particularly when analyzing limits or points of interest like where the segments meet, in this case, at \(x = 0\).

Exploring Continuity in Functions

Continuity in functions means the function has no breaks, jumps, or holes at a particular point. For a function to be continuous at a point, the limit approaching from the left must be equal to the value at the point, and it should match the limit approaching from the right.
To verify continuity for our piecewise function at \(x = 0\), we need:

• The left-hand limit, \(\lim_{x \to 0^{-}} f(x)\), to equal the right-hand limit, \(\lim_{x \to 0^{+}} f(x)\).
• The equal limit values must also correspond to the function value at that point, i.e., \(f(0)\).

In our example, both limits as \(x\) nears zero converge to zero. Moreover, given the function's defined pieces, the result \(\lim_{x \to 0} f(x) = 0\) suggests continuity at that point. Continuity proves invaluable for studying smooth transitions in calculus without sudden changes in diagram or function nature. It ensures equations are reliably behaving within their considered intervals.