Problem 74 - Calculus: Early Transcendentals

Question

One-sided limits Let $$f(x)=\left\\{\begin{array}{ll} 0 & \text { if } x \leq-5 \\ \sqrt{25-x^{2}} & \text { if }-5 < x < 5 \\ 3 x & \text { if } x \geq 5 \end{array}\right.$$ Compute the following limits or state that they do not exist. a. \(\lim _{x \rightarrow-5} f(x)\) b. \(\lim _{x \rightarrow-5^{+}} f(x)\) c. \(\lim _{x \rightarrow-5} f(x)\) d. \(\lim _{x \rightarrow 5^{-}} f(x)\) e. \(\lim _{x \rightarrow 5^{+}} f(x)\) f. \(\lim _{x \rightarrow 5} f(x)\)

Step-by-Step Solution

Answer

Answer: The limit of the given piecewise function as x approaches -5 is 0, and the limit as x approaches 5 does not exist.

Step 1: a. Determine the limit as x approaches -5

To find \(\lim_{x \rightarrow -5} f(x)\), we need to consider the definition of the function when \(x \leq -5\). In this case, \(f(x) = 0\). Since the function is constant in this interval, the limit exists and is equal to the constant value. Hence, \(\lim_{x \rightarrow -5} f(x) = 0\).

Step 2: b. Determine the limit as x approaches -5 from the right

To find the right-sided limit \(\lim_{x \rightarrow -5^{+}} f(x)\), we must consider the definition of the function when \(-5 < x < 5\). In this interval, \(f(x) = \sqrt{25-x^2}\). Since the function is continuous in this interval, then as x approaches -5 from the right, we can simply evaluate the function at x = -5: \(\lim_{x \rightarrow -5^{+}} f(x) = \sqrt{25 - (-5)^2} = 0\).

Step 3: c. Determine the limit as x approaches -5 (repeated)

This question repeats part a. So, the answer is the same: \(\lim_{x \rightarrow -5} f(x) = 0\).

Step 4: d. Determine the limit as x approaches 5 from the left

To find the left-sided limit \(\lim_{x \rightarrow 5^{-}} f(x)\), we must consider the definition of the function when \(-5 < x < 5\). In this interval, \(f(x) = \sqrt{25 - x^2}\). Since the function is continuous in this interval, then as x approaches 5 from the left, we can simply evaluate the function at x = 5: \(\lim_{x \rightarrow 5^{-}} f(x) = \sqrt{25 - (5)^2} = 0\).

Step 5: e. Determine the limit as x approaches 5 from the right

To find the right-sided limit \(\lim_{x \rightarrow 5^{+}} f(x)\), we must consider the definition of the function when \(x \geq 5\). In this interval, \(f(x) = 3x\). Since the function is continuous in this interval, then as x approaches 5 from the right, we can simply evaluate the function at x = 5: \(\lim_{x \rightarrow 5^{+}} f(x) = 3(5) = 15\).

Step 6: f. Determine the limit as x approaches 5

To find \(\lim_{x \rightarrow 5} f(x)\), we need to compare the left-sided and right-sided limits at x = 5. From our previous calculations, we have \(\lim_{x \rightarrow 5^{-}} f(x) = 0\) and \(\lim_{x \rightarrow 5^{+}} f(x) = 15\). Since the left-sided and right-sided limits are not equal, the limit at x = 5 does not exist. Hence, \(\lim_{x \rightarrow 5} f(x)\) does not exist.

Key Concepts

Limits of Piecewise Functions

Piecewise functions are defined by different expressions for different intervals of the domain. To understand limits of these functions, it is crucial to consider the specific interval relevant to the approaching point. For instance, in the given problem, the function f(x) is defined differently for different ranges of x, so the limits as x approaches -5 and 5 need to be assessed based on the definition of the function in those intervals.

When calculating the limit for a value located at an interval endpoint, like -5 or 5, we have to look for what happens as the variable x gets close to this number from the left and the right. This is where one-sided limits come into play, denoted as \( \lim_{x \rightarrow c^{-}} f(x) \) for the left-handed limit and \( \lim_{x \rightarrow c^{+}} f(x) \) for the right-handed limit. If these one-sided limits are not equal, as in the case of \( \lim_{x \rightarrow 5} f(x) \) in the exercise, the overall limit does not exist.

One-sided limits are particularly useful for dealing with discontinuities, where the function's behavior changes abruptly. The step by step solution carefully finds the value of the function as x approaches the transition points from appropriate directions, ensuring an accurate calculation of these limits.

Continuity and Limits

The concept of continuity is deeply intertwined with limits. A function is continuous at a point if the limit exists and is equal to the function's value at that point. Breaking it down, for continuity at point c, the function needs to satisfy three conditions: the function is defined at c, the limit as x approaches c exists, and this limit equals the function's value at c, written as \( \lim_{x \rightarrow c} f(x) = f(c) \).

In the context of the provided exercise, we can see that the function f(x) is not continuous at x = 5 because the left and right limits as x approaches 5 yield different values. The right-sided limit gives us 15, while the left-sided yields 0, and since they aren't equal, the function has a jump discontinuity here.

Understanding continuity is essential for predicting the behavior of functions at particular points and is also useful for establishing the existence of limits. The existing solution demonstrates precisely how changes in a piecewise function's definition can affect continuity, and why detailed examination at the boundaries is vital for providing a complete analysis.

Calculating Limits

Calculating limits, the fundamental process of determining the value that a function approaches as the input approaches a certain value, relies on understanding the behavior of the function near the point of interest. There are techniques such as direct substitution, factoring, rationalizing, and using special limit laws. However, in the case of piecewise functions, such as in our exercise, it's often a matter of assessing the relevant piece of the function near the point of interest and applying the correct approach.

In our exercise, direct substitution was possible for determining the one-sided limits since the function pieces were defined right up to the boundary points. For example, in step 2, since the function \( \sqrt{25 - x^2} \) is continuous and defined at the point approaching -5 from the right, we can directly substitute -5 into the function to find the limit.

When dealing with limits, it is also essential to consider special cases where direct substitution is not possible. For instance, indeterminate forms such as 0/0 may require additional techniques like factoring or applying L'Hôpital's Rule. However, these techniques were not needed in this exercise specifically because the limits involved could be calculated through simpler means. The step-by-step solution showcased the direct evaluation method, a fundamental tool for calculating limits in various contexts.