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Chapter 2: Problem 19

Find the limits in Exercises \(17-22\). $$ \begin{array}{ll}{\lim \left(\frac{x^{2}}{2}-\frac{1}{x}\right) \text { as }} \\\ {\begin{array}{ll}{\text { a. } x \rightarrow 0^{+}} & {\text { b. } x \rightarrow 0^{-}} \\ {\text { c. } x \rightarrow \sqrt[3]{2}} & {\text { d. } x \rightarrow-1}\end{array}}\end{array} $$

Short Answer

Expert verified
a: \(-\infty\), b: \(+\infty\), c: finite number, d: \(\frac{3}{2}\).

Step by step solution

01

Limit as x approaches 0 from the positive side

We are tasked to evaluate \( \lim_{{x \to 0^+}} \left( \frac{x^2}{2} - \frac{1}{x} \right) \). As \(x\) approaches 0 from the right, \( \frac{x^2}{2} \to \frac{0^2}{2} = 0 \) and \(-\frac{1}{x}\) approaches \(-\infty\) because \( \frac{1}{x} \) grows large negatively. Thus, the overall limit will be \(-\infty\).
02

Limit as x approaches 0 from the negative side

Now we consider \( \lim_{{x \to 0^-}} \left( \frac{x^2}{2} - \frac{1}{x} \right) \). As \(x\) approaches zero from the left, \( \frac{x^2}{2} \to 0 \) (as squares of small numbers are positive), and \(-\frac{1}{x}\) approaches \(+\infty\), because a negative \(x\) makes \( -\frac{1}{x} \) positive and large. Thus, the limit is \(+\infty\).
03

Limit as x approaches cube root of 2

Evaluate \( \lim_{{x \to \sqrt[3]{2}}} \left( \frac{x^2}{2} - \frac{1}{x} \right) \). Plugging \(x = \sqrt[3]{2}\) gives \( \frac{(\sqrt[3]{2})^2}{2} - \frac{1}{\sqrt[3]{2}} \). Calculate each part to get \(\frac{\sqrt[3]{4}}{2} - \sqrt[3]{\frac{1}{2}}\). This is a finite value since both terms are real numbers.
04

Limit as x approaches -1

Find \( \lim_{{x \to -1}} \left( \frac{x^2}{2} - \frac{1}{x} \right) \). Substitute \(x = -1\) to get \( \frac{(-1)^2}{2} - \frac{1}{-1} = \frac{1}{2} - (-1) = \frac{1}{2} + 1 = \frac{3}{2} \). As both parts are finite, the result is \( \frac{3}{2} \).

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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 help us understand the behavior of a function as it approaches a specific point, from one direction. When evaluating one-sided limits, we specify if we are approaching from the right \(x \rightarrow 0^+\) or from the left \(x \rightarrow 0^-\).

  • Right-hand limit: As \(x\rightarrow 0^+\), the expression \(\frac{x^2}{2} - \frac{1}{x}\) shows that \(\frac{x^2}{2}\) approaches zero, while \(-\frac{1}{x}\) becomes negatively infinite. Hence, the limit is \(-\infty\).

  • Left-hand limit: In contrast, when \(x \rightarrow 0^-\), \(\frac{x^2}{2}\) still approaches zero, but now \(-\frac{1}{x}\) approaches positive infinity as the division of a negative value becomes positive. Thus, the limit is \(+\infty\).
By comparing these, we can determine the general behavior of the function around the point of interest.
Infinite Limits
Infinite limits are used when a function grows without bound as it approaches a particular point. This concept occurs often with divisions by very small number values, as the divisor approaches zero.

When discussing \(\lim_{{x \to 0^+}} ( \frac{x^2}{2} - \frac{1}{x})\), the term \(-\frac{1}{x}\) becomes infinitely negative. This yields \(-\infty\) as the limit since the negative division approaches infinity. Meanwhile, for \(\lim_{{x \to 0^-}}\), \(-\frac{1}{x} \) transitions from negative to positive infinity and gives a result of \(+\infty\).
  • A function reaching a value of \(\pm \infty\) implies an asymptote might be present at that point.
  • Recognizing whether the function approaches infinity (positive or negative) is vital for understanding its extended behavior near those values.
Finite Limits
Finite limits occur when both parts of the function hold real and bounded values as they approach a given point. In such scenarios, we find particular numerical results rather than infinity.

For \(\lim_{{x \to \sqrt[3]{2}}} ( \frac{x^2}{2} - \frac{1}{x})\), substituting \(x = \sqrt[3]{2}\), the expression simplifies to \( \frac{\sqrt[3]{4}}{2} - \sqrt[3]{\frac{1}{2}}\). Here, both components are real numbers, rendering a calculable and finite outcome.

Another example is when \(x\rightarrow-1\), yielding \(\lim_{{x \to -1}} ( \frac{x^2}{2} - \frac{1}{x}) = \frac{3}{2}\). The finite limit exists since both the squared division and linear terms result in finite numbers.
  • Finite limits indicate that a function stabilizes around specific values.
  • They help in identifying smoother behavior as a function approaches the point.
Limit Evaluation Techniques
To evaluate limits effectively, several approaches can be used, especially when dealing with complex expressions.

  • Direct Substitution: This technique involves plugging in the values directly. If the function is continuous, like as \(x\rightarrow -1\), solves easily this way.

  • Simplification: Simplify equations whenever values can't be directly substituted. For instance, when dealing with cube roots and fractions.

  • Understanding Infinity Behavior: Look at the growth of expressions as they head toward infinite sizes, such as observing \( \frac{1}{x} \).
These techniques, employed correctly, allow for precise calculations and a better grasp of the function's behavior around the point of interest. They guide students in predicting how the expressions behave near different points.

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Most popular questions from this chapter

A function discontinuous at every point a. Use the fact that every nonempty interval of real numbers contains both rational and irrational numbers to show that the function $$ f(x)=\left\\{\begin{array}{ll}{1,} & {\text { if } x \text { is rational }} \\\ {0,} & {\text { if } x \text { is rrational }} \\ {0,} & {\text { if } x \text { is irrational }}\end{array}\right. $$ is discontinuous at every point. b. Is \(f\) right-continuous or left-continuous at any point?

Define \(f(1)\) in a way that extends \(f(s)=\left(s^{3}-1\right) /\left(s^{2}-1\right)\) to be continuous at \(s=1 .\)

In Exercises \(19-22,\) find the slope of the curve at the point indicated. $$ y=\frac{1}{x-1}, \quad x=3 $$

Graphing Secant and Tangent Lines Use a CAS to perform the following steps for the functions in Exercises \(45-48 .\) a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\) b. Holding \(x_{0}\) fixed, the difference quotient $$ q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h} $$ at \(x_{0}\) becomes a function of the step size \(h .\) Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\) d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h=3,2\) and \(1 .\) Graph them together with \(f\) and the tangent line over the interval in part (a). $$ f(x)=x+\sin (2 x), \quad x_{0}=\pi / 2 $$

In Exercises \(61-66,\) you will further explore finding deltas graphically. Use a CAS to perform the following steps: a. Plot the function \(y=f(x)\) near the point \(x_{0}\) being approached. b. Guess the value of the limit \(L\) and then evaluate the limit symbolically to see if you guessed correctly. c. Using the value \(\epsilon=0.2,\) graph the banding lines \(y_{1}=L-\epsilon\) and \(y_{2}=L+\epsilon\) together with the function \(f\) near \(x_{0}\) . d. From your graph in part (c), estimate a \(\delta > 0\) such that for all \(x\) $$ 0 < \left|x-x_{0}\right| < \delta \quad \Rightarrow \quad|f(x)-L| < \epsilon $$ Test your estimate by plotting \(f, y_{1},\) and \(y_{2}\) over the interval \(0 < \left|x-x_{0}\right| < \delta .\) For your viewing window use \(x_{0}-2 \delta \leq x \leq x_{0}+2 \delta\) and \(L-2 \epsilon \leq y \leq L+2 \epsilon\) . If any function values lie outside the interval \([L-\epsilon, L+\epsilon],\) your choice of \(\delta\) was too large. Try again with a smaller estimate. e. Repeat parts (c) and (d) successively for \(\epsilon=0.1,0.05,\) and \(0.001 .\) $$ f(x)=\frac{3 x^{2}-(7 x+1) \sqrt{x}+5}{x-1}, \quad x_{0}=1 $$
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