Problem 34 - Calculus Early Transcendentals

Question

Let \(f(x)=\left\\{\begin{array}{ll}0 & \text { if } x \leq-5 \\\ \sqrt{25-x^{2}} & \text { if }-5

Step-by-Step Solution

Answer

Question: Determine the limits of the piecewise function \(f(x)\) as \(x\) approaches -5 and 5. Answer: The limit as \(x\) approaches -5 is 0 because both the left and the right limits are equal at 0: \(\lim_{x \rightarrow -5}f(x) = 0\). The limit as \(x\) approaches 5 does not exist because the left and the right limits are not equal (0 and 15, respectively): \(\lim_{x \rightarrow 5}f(x)\) does not exist.

Step 1: Limit a: \(\lim_{x \rightarrow -5^-}f(x)\)

Since \(x\) is approaching -5 from the left, we should look at the part of the function that is defined for \(x \leq -5\): \(f(x) = 0\). The limit as \(x\) approaches -5 from the left is simply 0.

Step 2: Limit b: \(\lim_{x \rightarrow -5^+}f(x)\)

Since \(x\) is approaching -5 from the right, we should look at the part of the function that is defined for \(-5 < x < 5\): \(f(x) = \sqrt{25-x^2}\). We can substitute -5 into this expression: \(\sqrt{25-(-5)^2} = \sqrt{0} = 0\). Therefore, the limit as \(x\) approaches -5 from the right is also 0.

Step 3: Limit c: \(\lim_{x \rightarrow -5}f(x)\)

Since both the left and the right limits as \(x\) approaches -5 are equal (both 0), the overall limit exists and is also 0: \(\lim_{x \rightarrow -5}f(x) = 0\).

Step 4: Limit d: \(\lim_{x \rightarrow 5^-}f(x)\)

Since \(x\) is approaching 5 from the left, we should look at the part of the function that is defined for \(-5 < x < 5\): \(f(x) = \sqrt{25-x^2}\). We can substitute 5 into this expression: \(\sqrt{25-5^2} = \sqrt{0} = 0\). Therefore, the limit as \(x\) approaches 5 from the left is 0.

Step 5: Limit e: \(\lim_{x \rightarrow 5^+}f(x)\)

Since \(x\) is approaching 5 from the right, we should look at the part of the function that is defined for \(x \geq 5\): \(f(x) = 3x\). We can substitute 5 into this expression to get \(3(5) = 15\). Therefore, the limit as \(x\) approaches 5 from the right is 15.

Step 6: Limit f: \(\lim_{x \rightarrow 5}f(x)\)

Since the left and the right limits as \(x\) approaches 5 are not equal (0 and 15, respectively), the overall limit does not exist: \(\lim_{x \rightarrow 5}f(x)\) does not exist.

Key Concepts

Piecewise functions

Piecewise functions are special types of functions that have different expressions based on the value of the variable. They are defined in pieces, which work like different segments or parts of the function. Each segment has its own specific rule or formula that applies to a specific range of the input values.
For example, in our exercise, the function \( f(x) \) is piecewise and has three distinct segments:

• When \( x \leq -5 \), the function is constant and equals 0.
• When \(-5 < x < 5\), the function is \( \sqrt{25-x^2} \).
• When \( x \geq 5 \), the function is a simple linear function \( 3x \).

This way, piecewise functions allow us to describe complex behaviors using simple, understandable rules. These types of functions are quite common in real-world problems where a situation or condition changes over different intervals.

Left-hand limit

The left-hand limit refers to the value that a function approaches as the input approaches a specific point from the left side. It's denoted as \( \lim_{x \to c^-} f(x) \), where \( c \) is the point of interest. To find the left-hand limit, we should consider only the segment of the piecewise function that applies for values less than or equal to the point.
In the example problem:

• For \( x = -5 \), the left-hand limit of \( f(x) \) is 0 because the function is defined as 0 for \( x \leq -5 \).

For another part of the problem:

• For \( x = 5 \), we check \( -5 < x < 5 \), substituting yields 0, making the left-hand limit 0.

Left-hand limits help us understand how a function behaves just before reaching a certain point, which is important in ensuring consistency in the behavior of functions.

Right-hand limit

A right-hand limit looks at the behavior of a function as the input approaches a point from the right. It is represented as \( \lim_{x \to c^+} f(x) \). To determine the right-hand limit, consider the expression from the piecewise function that applies to values greater than the point of interest.
In our example:

• For \( x = -5 \), the right-hand limit of \( f(x) \) is 0, using the expression \( \sqrt{25-x^2} \) for \(-5 < x < 5\) and substituting \( x = -5 \).
• For \( x = 5 \), we use the formula \( 3x \) for \( x \geq 5 \), resulting in \( 15 \).

Right-hand limits reveal the value that a function approaches as we come from the right of a point. This is crucial when analyzing functions for continuity or points where behavior changes.

Overall limit

The overall limit at a point exists if and only if both the left-hand and right-hand limits at that point are equal. This is represented as \( \lim_{x \to c} f(x) \). If the left-hand and right-hand limits differ at a point, it means there is a jump or discontinuity at that point, and hence, the overall limit does not exist.
In the exercise example:

• At \( x = -5 \), both the left-hand and right-hand limits are 0. Therefore, the overall limit is 0.
• At \( x = 5 \), the left-hand limit is 0 and the right-hand limit is 15, indicating a discontinuity. Hence, the overall limit does not exist at this point.

Understanding overall limits helps to classify points of continuity or discontinuity in piecewise functions, which is essential for analyzing the behavior of such functions fully.