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Magazine Chapter 2: Problem 55
Find the limits. \(\lim \left(\frac{x^{2}}{2}-\frac{1}{x}\right)\) as a. \(\quad x \rightarrow 0^{+}\) b. \(x \rightarrow 0^{-}\) c. \(x \rightarrow \sqrt[3]{2}\) d. \(x \rightarrow-1\)
Short Answer
Expert verified
a. \(-\infty\); b. \(+\infty\); c. \(\frac{\sqrt[3]{4}}{2} - \frac{1}{\sqrt[3]{2}}\); d. \(\frac{3}{2}\).
Step by step solution
01
Identify the Function
The function given is \(f(x) = \frac{x^{2}}{2} - \frac{1}{x}\). We need to find the limit of this function as \(x\) approaches the different given values.
02
Limit as \(x \rightarrow 0^{+}\)
To find the limit as \(x\) approaches \(0^{+}\), substitute \(x\) values that are positive and very close to zero into the function. The term \(-\frac{1}{x}\) dominates because it tends to \(-\infty\) as \(x\) goes to zero from the positive side, leading to: \[\lim_{{x \to 0^{+}}}\left(\frac{x^{2}}{2} - \frac{1}{x}\right) = -\infty.\]
03
Limit as \(x \rightarrow 0^{-}\)
Now, to find the limit as \(x\) approaches \(0^{-}\), substitute \(x\) values that are negative and very close to zero into the function. The term \(-\frac{1}{x}\) again dominates because it tends to \(+\infty\) as \(x\) goes to zero from the negative side, leading to: \[\lim_{{x \to 0^{-}}}\left(\frac{x^{2}}{2} - \frac{1}{x}\right) = +\infty.\]
04
Limit as \(x \rightarrow \sqrt[3]{2}\)
Here, substitute \(x = \sqrt[3]{2}\) into the function: \[f(x) = \frac{(\sqrt[3]{2})^2}{2} - \frac{1}{\sqrt[3]{2}} = \frac{\sqrt[3]{4}}{2} - \frac{1}{\sqrt[3]{2}}.\]As \(x\) approaches \(\sqrt[3]{2}\), both terms are well-defined and finite:\[\lim_{{x \to \sqrt[3]{2}}} \left(\frac{x^{2}}{2} - \frac{1}{x}\right) = \frac{\sqrt[3]{4}}{2} - \frac{1}{\sqrt[3]{2}}.\]
05
Limit as \(x \rightarrow -1\)
Now, evaluate the limit as \(x\) approaches \(-1\). Substitute \(x = -1\) into the function:\[f(x) = \frac{(-1)^2}{2} - \frac{1}{-1} = \frac{1}{2} + 1 = \frac{3}{2}.\]The direct substitution shows the limit is:\[\lim_{{x \to -1}} \left(\frac{x^{2}}{2} - \frac{1}{x}\right) = \frac{3}{2}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
One-Sided Limits
In calculus, one-sided limits are a fundamental concept that deals with the behavior of a function as the input approaches a particular point from one side鈥攅ither from the left or from the right. They help determine the limit of a function as it gets closer to the value from one direction only.
In our exercise, we observe both right-hand limits and left-hand limits at the point where the function approaches zero.
In our exercise, we observe both right-hand limits and left-hand limits at the point where the function approaches zero.
- For the limit as \(x \to 0^{+}\), we're considering values of \(x\) that are greater than zero but very close to zero. Here, the term \(-\frac{1}{x}\) becomes significantly negative, leading the overall function towards \(-\infty\).
- For the limit as \(x \to 0^{-}\), we're looking at values less than zero nearing zero. In this scenario, the term \(-\frac{1}{x}\) becomes large and positive, pushing the function towards \(+\infty\).
Infinite Limits
Infinite limits occur when a function approaches an infinitely large positive or negative value as the input approaches a certain point. This often signals a vertical asymptote鈥攁 line that the graph of a function gets very close to, but never touches.
In our problem, the function itself suggests infinite behavior near zero due to the \(-\frac{1}{x}\) component.
In our problem, the function itself suggests infinite behavior near zero due to the \(-\frac{1}{x}\) component.
- As \(x\) approaches 0 from the positive side, the reciprocal term \(-\frac{1}{x}\) grows negatively without bound, resulting in \(-\infty\).
- Conversely, as \(x\) nears 0 from the negative side, this term grows positively without limit, giving \(+\infty\).
Evaluating Limits
Evaluating limits is the process of finding the value that a function approaches as the input nears a specific point. It's important in determining continuity and solving real-world problems involving change and motion.
In our solution, we evaluated limits at different points:
In our solution, we evaluated limits at different points:
- For \(x \to \sqrt[3]{2}\), by directly substituting \(x\) with \(\sqrt[3]{2}\), the terms provide a concrete finite value due to non-zero denominator and well-defined components. Hence, the limit is evaluated as \(\frac{\sqrt[3]{4}}{2} - \frac{1}{\sqrt[3]{2}}\).
- Similarly, approaching \(x = -1\) involves direct substitution, yielding the function value \(\frac{3}{2}\).
Function Behavior Near Point
Understanding how a function behaves near a particular point involves examining its limits at that point, assisting in understanding continuity and identifying discontinuities.
In this example:
In this example:
- Near \(x = 0\), the function behavior is highlighted by different one-sided limits, indicating a discontinuity there, and necessitates separate evaluations on either side of 0.
- At \(x = \sqrt[3]{2}\) and \(x = -1\), the function is continuous, and does not require one-sided limit consideration as direct substitution provides a value, indicating the smoothness of the graph at those points.
