Problem 31 - Single Variable Calculus

Question

Let $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{x} & \text { if } x<0 \\ \sin x & \text { if } 0 \leq x<\pi \\ 0 & \text { if } x \geq \pi \end{array}\right. $$ a. Sketch the graph of \(f\). b. Find all values of \(x\) in the domain of \(f\) at which the limit of \(f\) exists. c. Find all values of \(x\) in the domain of \(f\) at which the lefthand limit of \(f\) exists. d. Find all values of \(x\) in the domain of \(f\) at which the righthand limit of \(f\) exists.

Step-by-Step Solution

Answer

The graph of \(f(x)\) consists of a hyperbola for \(x<0\), a sine function for \(0 \le x < \pi\), and a horizontal line for \(x \ge \pi\). The limit of \(f(x)\) exists for all values of \(x\) in the domain. The left-hand limit and right-hand limit of \(f(x)\) also exist for all values of \(x\) in the domain.

Step 1: Sketch the graph of the function

For \(x < 0\), we are given \(f(x) = \frac{1}{x}\). This part of the function represents a hyperbola. For \(0 \leq x < \pi\), we have \(f(x) = \sin x\), which is a sine function, and for \(x \geq \pi\), we have \(f(x) = 0\), representing a horizontal line. By combining all three parts of the function, we can generate a sketch of the graph as follows: 1. The hyperbola part: Draw the graph of \(y = \frac{1}{x}\) for \(x<0\) 2. The sine part: Connect the sine function to the hyperbola at the origin, draw the graph of \(y = \sin x\) for the interval \(0 \le x < \pi\) 3. The horizontal line part: Draw the graph of \(y = 0\) starting at \(x = \pi\) We should have a coherent sketch of the piecewise function that incorporates all three parts.

Step 2: Identify where the limit of f exists

A limit exists at a point \(x\) if both the left-hand limit and the right-hand limit exist at that point and are equal. In this case, we can identify that the limit exists for all \(x < 0\), since \(f(x) = \frac{1}{x}\) and \(\lim_{x \to a} \frac{1}{x}\) exists for all a in this domain. The limit also exists for all values of \(x\) in the interval \(0 \le x < \pi\), since the sine function is continuous in this domain. Lastly, as the horizontal line at \(f(x) = 0\) represents a limit of zero for every point where \(x \ge \pi\), we can conclude that the limit exists for all values of \(x \ge \pi\) as well. Thus, the limit of \(f(x)\) exists for all values of \(x\) in the domain of the function.

Step 3: Identify values where the left-hand limit of f exists

To find the left-hand limit, we need to determine if the limit exists when the function is approached from the left, i.e., when \(x\) is slightly smaller than the given value. The left-hand limit exists for all \(x < 0\) since the limit of \(f(x) = \frac{1}{x}\) exists when approached from the left for all values in this domain. For \(0\le x < \pi\), the left-hand limit exists as well, because it's continuous in this domain. Finally, for \(x \ge \pi\), the left-hand limit exists since the horizontal line at \(f(x) = 0\) has a constant value and as such also has a left-hand limit for \(x \ge \pi\). Hence, the left-hand limit of \(f(x)\) exists for all values of \(x\) in the domain of the function.

Step 4: Identify where the right-hand limit of f exists

Now to find the right-hand limit, we need to determine if the limit exists when the function is approached from the right, i.e., when \(x\) is slightly greater than the given value. For \(x < 0\), the right-hand limit exists because the limit of \(f(x) = \frac{1}{x}\) exists when approached from the right for all values in this domain. For \(0 \le x < \pi\), the right-hand limit exists since the function is continuous in this domain, indicating the existence of the right-hand limit. Lastly, for \(x \ge \pi\), the right-hand limit exists since the horizontal line at \(f(x) = 0\) has a constant value and also has a right-hand limit for \(x \ge \pi\). Therefore, the right-hand limit of \(f(x)\) exists for all values of \(x\) in the domain of the function.

Key Concepts

Limits

When discussing limits, we're exploring what happens to a function as it approaches a particular value. In simpler terms, we're trying to understand the behavior of a function as it gets close to a specific point. For our piecewise function, the question is all about where the limits exist.

• For the part of the graph defined by \(f(x) = \frac{1}{x}\) for \(x < 0\), the function is continuous and well-behaved, so the limits exist wherever \(x < 0\).
• In the segment \(0 \leq x < \pi\), we have \(f(x) = \sin x\). Here, the limit exists because the sine function is inherently continuous, which means it flows smoothly without any breaks or gaps.
• Finally, for \(x \geq \pi\), \(f(x) = 0\). This part of the function behaves like a constant horizontal line, so the limit is continuous across the places where this section applies.

In summary, the limits of the function exist over the entire domain. This means, at any point on the function, both what comes from the left and right meet where expected, giving us predictable behavior everywhere we're allowed to look at the function.

Continuity

Continuity in functions means you can draw them without lifting your pencil. In our exercise, we're looking to see where the piecewise function is continuous, which involves walking through each piece of the function.

• For \(x < 0\), continuity is present because \(f(x) = \frac{1}{x}\) doesn't have any interruptions. The hyperbola continues smoothly.
• In the range \(0 \leq x < \pi\), with \(f(x) = \sin x\), continuity holds since the sine curve naturally transitions without jumps or surprises. The whole interval gives you a nice smooth wave.
• For \(x \geq \pi\), \(f(x)\) being zero means there's a straight flat line, showing continuity as there's no change as you move along the x-axis.

It’s important to note how each section fits together. Specifically, the function may have a discontinuity at \(x = 0\), which is where the pieces shift from the hyperbolic part to the sine function. The potential discontinuity at \(x = 0\) is remedied because as \(x\) approaches zero from the left, \(f(x)\) does not meet the value of the sine function starting at zero. But within their respective parts, all segments are continuous.

Graph Sketching

Sketching the graph of a piecewise function can feel like piecing together parts of different puzzles. You will need to draw each section of the piece appropriately while ensuring they align correctly at any points where the function 'switches.'

To begin:

• Sketch \(f(x) = \frac{1}{x}\) for \(x < 0\). This will give you a hyperbola in the left portion of your graph. Remember, with \(x o 0^-\), \(f(x)\) tends to negative infinity, indicating how sharp this hyperbola approaches the y-axis near zero.
• For \(0 \leq x < \pi\), sketch the sine wave \(f(x) = \sin x\). Start from the origin and smoothly draw the wave, peaking at one and falling back towards zero near \(x = \pi\).
• When \(x \geq \pi\), draw the horizontal line where \(f(x) = 0\). This carries on past \(x = \pi\), an enduring flat line across the remaining portion of the graph. Make sure this part of the graph doesn't connect but starts exactly at \(x = \pi\).

To ensure accuracy, notice how well these parts fit together at each transitional boundary. These transitions tell you where the graph will exhibit distinct features, like possible discontinuities or smooth connections.