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Find the limits. a. \(\lim _{x \rightarrow 0^{+}} \frac{2}{x^{1 / 5}}\) b. \(\lim _{x \rightarrow 0^{-}} \frac{2}{x^{1 / 5}}\)

Short Answer

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

Step by step solution

01

Understanding the problem

We need to determine the behavior of the function \( f(x) = \frac{2}{x^{1/5}} \) as \( x \) approaches 0 from the positive side for part (a) and from the negative side for part (b). The fifth root function \( x^{1/5} \) behaves differently depending if \( x \) is positive or negative.
02

Analyze \\ x^{1/5} \\ as \\ x \\ approaches \\ 0^{+}

As \( x \) approaches \( 0^{+} \), \( x^{1/5} \) also approaches \( 0^{+} \) since the fifth root of a positive number is positive. Thus, \( \frac{2}{x^{1/5}} \) will become \( \frac{2}{0^{+}} \), which suggests the limit tends to \( +\infty \).
03

Solve part (a)

Based on the analysis in Step 2, calculate the limit: \[ \lim_{x \to 0^{+}} \frac{2}{x^{1/5}} = +\infty \] because the denominator approaches 0 from the positive side, making the fraction diverge towards positive infinity.
04

Analyze \\ x^{1/5} \\ as \\ x \\ approaches \\ 0^{-}

As \( x \) approaches \( 0^{-} \), \( x^{1/5} \) approaches a small negative number because the fifth root of a negative number is also negative. Hence, \( \frac{2}{x^{1/5}} \) will become \( \frac{2}{0^{-}} \), which suggests the limit tends to \( -\infty \).
05

Solve part (b)

Based on the analysis in Step 4, calculate the limit: \[ \lim_{x \to 0^{-}} \frac{2}{x^{1/5}} = -\infty \] because the denominator approaches 0 from the negative side, making the fraction diverge towards negative infinity.

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Key Concepts

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

Understanding One-Sided Limits
In calculus, sometimes we are interested in understanding how a function behaves as it approaches a specific point from one side only. This is known as a one-sided limit. We have the limit from the right, often written as \( \lim_{x \to a^+} f(x) \), and the limit from the left, \( \lim_{x \to a^-} f(x) \). These symbols indicate we are exploring the behavior of the function as \( x \) gets closer to \( a \) from values larger than \( a \), or smaller than \( a \), respectively.
In the given exercise, we found two one-sided limits for the function \( f(x) = \frac{2}{x^{1/5}} \) as \( x \) approaches 0 both from positive and negative directions.

For the limit from the positive side, as \( x \to 0^+ \), the denominator becomes very small, leading to an infinitely large positive value, hence the limit is \( +\infty \).
Conversely, when \( x \to 0^- \), the denominator being a small negative number causes the fraction to tend to a large negative value, thus \( -\infty \).
  • Right-hand limits look at approaching values from larger numbers.
  • Left-hand limits examine from smaller numbers.
  • These help understand sharp changes or undefined points.
Exploring Infinite Limits
Infinite limits describe situations where a function's value increases or decreases without bound as it approaches a specific point. An infinite limit doesn't give a finite number but rather shows the function's behavior as tending to infinity.
In the problem, we explored the limits of \( f(x) = \frac{2}{x^{1/5}} \) as \( x \) approaches 0. Here, because the denominator approaches 0, the overall fraction grows or shrinks without limit depending on the direction from which \( x \) approaches zero.

These infinite limits can help us understand vertical asymptotes in a function's graph. A vertical asymptote occurs where the function reaches exceedingly large values, much like \( x = 0 \) for the given function.
  • Infinite limits approach large magnitudes.
  • They indicate vertical asymptotes or sharp changes.
  • Essential in analyzing function behavior around critical points.
Behavior of Functions Near a Point
Examining how functions behave as they near a particular point is crucial in calculus. This analysis can uncover much about the nature of the function at that point, including rates of change and potential discontinuities.
The exercise focused on understanding the behavior as \( x \) nears zero for the function \( f(x) = \frac{2}{x^{1/5}} \). The fifth root function behaves uniquely:
  • As \( x \to 0^+ \), \( x^{1/5} \to 0^+ \), leading \( f(x) \) to grow towards \( +\infty \). This shows a sharp vertical asymptote.
  • As \( x \to 0^- \), \( x^{1/5} \) also approaches 0 but is negative, thus \( f(x) \to -\infty \), indicating an opposite vertical asymptote.
  • This differing behavior from each side reflects a point of discontinuity.
Understanding how a function changes around critical points can help in designing graphs and predicting potential real-world applications where sudden changes occur.

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Most popular questions from this chapter

You will find a graphing calculator useful for Exercise. Let \(f(x)=\left(x^{2}-9\right) /(x+3).\) a. Make a table of the values of \(f\) at the points \(x=-3.1\) \(-3.01,-3.001,\) and so on as far as your calculator can go. Then estimate \(\lim _{x \rightarrow-3} f(x) .\) What estimate do you arrive at if you evaluate \(f\) at \(x=-2.9,-2.99,-2.999, \ldots\) instead? b. Support your conclusions in part (a) by graphing \(f\) near \(c=-3\) and using Zoom and Trace to estimate \(y\) -values on the graph as \(x \rightarrow-3.\) c. Find \(\lim _{x \rightarrow-3} f(x)\) algebraically, as in Example 7.

Let \(f\) be defined on an interval \((a, b)\) and suppose that \(f(c) \neq 0\) at some \(c\) where \(f\) is continuous. Show that there is an interval \((c-\delta, c+\delta)\) about \(c\) where \(f\) has the same sign as \(f(c)\)

Use a CAS to perform the following steps: a. Plot the function \(y=f(x)\) near the point \(c\) 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 \(c\) d. From your graph in part (c), estimate a \(\delta>0\) such that for all \(x\) \(0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon\) Test your estimate by plotting \(f, y_{1},\) and \(y_{2}\) over the interval \(0<|x-c|<\delta .\) For your viewing window use \(c-2 \delta \leq\) \(x \leq c+2 \delta\) 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 $$f(x)=\frac{x(1-\cos x)}{x-\sin x}, \quad c=0$$

Show that the function \(F(x)=(x-a)^{2}\) \((x-b)^{2}+x\) takes on the value \((a+b) / 2\) for some value of \(x\)

You will find a graphing calculator useful for Exercise. Let \(g(x)=\left(x^{2}-2\right) /(x-\sqrt{2}).\) a. Make a table of the values of \(g\) at the points \(x=1.4,1.41\) 1.414, and so on through successive decimal approximations of \(\sqrt{2} .\) Estimate \(\lim _{x \rightarrow \sqrt{2}} g(x).\) b. Support your conclusion in part (a) by graphing \(g\) near \(c=\sqrt{2}\) and using Zoom and Trace to estimate \(y\) -values on the graph as \(x \rightarrow \sqrt{2}.\) c. Find \(\lim _{x \rightarrow \sqrt{2}} g(x)\) algebraically.

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