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Chapter 2: Problem 59

Find the limits. \(\lim \left(2-\frac{3}{t^{1 / 3}}\right) \mathrm{as}\) a. \(t \rightarrow 0^{+}\) b. \(t \rightarrow 0^{-}\)

Short Answer

Expert verified
a. \(-\infty\) b. \(+\infty\)

Step by step solution

01

Understanding the Problem

We need to find the limits of the function \( f(t) = 2 - \frac{3}{t^{1/3}} \) as \( t \) approaches zero from the positive side (\( t \to 0^+ \)), and from the negative side (\( t \to 0^- \)). This involves analyzing the behavior of the cube root term \( t^{1/3} \) when \( t \) is close to zero.
02

Analyzing Limit as t Approaches Zero from the Positive Side

When \( t \to 0^+ \), \( t^{1/3} \) is a small positive number. In this scenario, \( \frac{3}{t^{1/3}} \) becomes very large and positive, causing \( 2 - \frac{3}{t^{1/3}} \) to approach \(-\infty\). Thus, the limit is \(-\infty\).
03

Analyzing Limit as t Approaches Zero from the Negative Side

When \( t \to 0^- \), \( t^{1/3} \) becomes a small negative number. As a result, \( \frac{3}{t^{1/3}} \) becomes very large but negative, leading \( 2 - \frac{3}{t^{1/3}} \) to approach \(+\infty\). Hence, the limit is \(+\infty\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Behavior of cube roots
The cube root function presents a unique behavior that sets it apart from square roots. Let's explore how this behavior impacts our limit problem. When dealing with cube roots, the function \( t^{1/3} \) represents the cube root of \( t \). This function has a consistent behavior whether \( t \) is positive or negative, which is not the case for square roots.
  • If \( t \) is positive, \( t^{1/3} \) is also positive, and shrinking \( t \) towards zero results in \( t^{1/3} \) becoming a smaller positive number.
  • If \( t \) is negative, \( t^{1/3} \) becomes negative, ensuring consistent behavior asymmetrically along the negative axis. In this scenario, shrinking \( t \) towards zero results in \( t^{1/3} \) being a small negative value.
This functionality is crucial in our limit analysis as it allows the division in \( \frac{3}{t^{1/3}} \) to behave differently when \( t \) approaches zero from positive and negative directions. The behavior of cube roots establishes the groundwork for understanding the subsequent impact on limits.
Approaching zero
Analyzing limits often involves understanding how a function behaves as its variable gets infinitesimally close to a specific value, in this case, zero. The expression \( \frac{3}{t^{1/3}} \) in the function stands out when \( t \) approaches zero. This is due to the influence of the denominator becoming a very tiny number.
  • As \( t \to 0^+ \), we're looking at very small positive numbers. These numbers make \( t^{1/3} \) shrink into smaller positives, causing \( \frac{3}{t^{1/3}} \) to skyrocket to large positive values.
  • Conversely, for \( t \to 0^- \), where \( t \) is a small negative, \( t^{1/3} \) also becomes a small negative. This flips \( \frac{3}{t^{1/3}} \) to large negative values because dividing by a tiny negative produces a large negative outcome.
Approaching zero with these minute variations plays a crucial role in determining if the output of the entire function heads towards positive or negative infinity. It embodies the unpredictable nature of limits around certain points.
Positive and negative limits
Limits can reveal different outcomes based on the direction from which the variable approaches a point. In this exercise, we see intriguing results depending on whether \( t \) approaches zero from the positive side or the negative side.
  • When \( t \to 0^+ \), the phenomenon where very small positive cube roots lead \( \frac{3}{t^{1/3}} \) to a large positive means that \( 2 - \frac{3}{t^{1/3}} \) shoots towards \(-\infty\). This leads to the conclusion that the limit value is negative infinity.
  • In contrast, with \( t \to 0^- \), the small negative cube roots turn \( \frac{3}{t^{1/3}} \) extremely negative, forcing \( 2 - \text{large negative} \) to move towards \(+\infty\). Here, the limit is positive infinity.
This dual behavior reflects the importance of understanding the direction of approach in determining limits. It showcases how function behavior can drastically change based on subtle shifts in input values, providing a deeper insight into calculus' intricate nature. Differentiating between positive and negative limits offers a comprehensive understanding of how functions can diverge to different infinities.

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Most popular questions from this chapter

a. Graph \(h(x)=x^{2} \cos \left(1 / x^{3}\right)\) to estimate \(\lim _{x \rightarrow 0} h(x),\) zooming in on the origin as necessary. b. Confirm your estimate in part (a) with a proof.

Graph the function \(f\) to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function's value at \(x=0 .\) If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function's value(s) should be? $$f(x)=\frac{\sin x}{|x|}$$

Gives a function \(f(x),\) a point \(c,\) and a positive number \(\epsilon .\) Find \(L=\lim f(x) .\) Then find a number \(\delta>0\) such that for all \(x\) $$0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon$$ $$f(x)=-3 x-2, \quad c=-1, \quad \epsilon=0.03$$

$$\text { Prove that } \lim _{x \rightarrow c} f(x)=L \text { if and only if } \lim _{h \rightarrow 0} f(h+c)=L$$

Manufacturing electrical resistors Ohm's law for electrical circuits like the one shown in the accompanying figure states that \(V=R I .\) In this equation, \(V\) is a constant voltage, \(I\) is the current in amperes, and \(R\) is the resistance in ohms. Your firm has been asked to supply the resistors for a circuit in which \(V\) will be 120 volts and \(I\) is to be \(5 \pm 0.1\) amp. In what interval does \(R\) have to lie for \(I\) to be within 0.1 amp of the value \(I_{0}=5 ?\) GRAPH CANT COPY Showing \(L\) is not a limit We can prove that \(\lim _{x \rightarrow c} f(x) \neq L\) by providing an \(\epsilon>0\) such that no possible \(\delta>0\) satisfies the condition $$\text { for all } x, \quad 0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon $$We accomplish this for our candidate \(\epsilon\) by showing that for each \(\delta>0\) there exists a value of \(x\) such that$$0<|x-c|<\delta \quad \text { and } \quad|f(x)-L| \geq \epsilon$$ GRAPH CANT COPY
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