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Chapter 2: Problem 29

Location of vertical asymptotes Analyze the following limits and find the vertical asymptotes of \(f(x)=\frac{x-5}{x^{2}-25}.\) \(\begin{array}{llll}\text { a. } \lim _{x \rightarrow 5} f(x) & \text { b. } \lim _{x \rightarrow-5^{-}} f(x) & \text { c. } \lim _{x \rightarrow-5^{+}} f(x)\end{array}\)

Short Answer

Expert verified
Answer: The vertical asymptote of the function is at \(x=-5\).

Step by step solution

01

Simplify the function

Firstly, we need to simplify the function. The denominator has a difference of squares which can be factored as \((x-5)(x+5)\), resulting in: \(f(x) = \frac{x-5}{(x-5)(x+5)}\) Now, we can cancel out the common factor \((x-5)\) from the numerator and the denominator: \(f(x) = \frac{1}{x+5}\) for \(x \neq 5\) It must be noted that the original function has a hole at \(x=5\), but it is not an asymptote, as the function is continuous at \(x=5\).
02

Determine the points where the denominator goes to 0

Now, we need to determine when the denominator of the simplified function goes to 0: \(x+5=0 \Rightarrow x=-5\) This shows that there is a vertical asymptote at \(x=-5\).
03

Analyze the limits for x approaching -5

Now, we need to analyze the limits given in the exercise for both left and right of the vertical asymptote at x=-5: a. \(\lim_{x \rightarrow 5} f(x)\) Since there is no vertical asymptote at x=5, as we already found it was a hole, we can simply evaluate the limit by plugging 5 into the simplified function: \(\lim_{x \rightarrow 5} \frac{1}{x+5} = \frac{1}{5+5} = \frac{1}{10}\) b. \(\lim_{ x \rightarrow -5^{-} } f(x)\) In this case, we are approaching -5 from the left. Let's plug in the simplified function: \(\lim_{ x \rightarrow -5^{-} } \frac{1}{x+5}\) As x approaches -5 from the left, the denominator goes to 0, and we have \(\frac{1}{-}(a small number)\), so the limit is \(-\infty\). c. \(\lim_{ x \rightarrow -5^{+} } f(x)\) Similarly, we are approaching -5 from the right: \(\lim_{ x \rightarrow -5^{+} } \frac{1}{x+5}\) As x approaches -5 from the right, the denominator goes to 0, but this time we have \(\frac{1}{+}(a tiny number)\), so the limit is \(+\infty\).
04

Conclusion

From our analysis, we have concluded the following: a. \(\lim_{x \rightarrow 5} f(x) = \frac{1}{10}\) b. \(\lim_{ x \rightarrow -5^{-} } f(x) = -\infty\) c. \(\lim_{ x \rightarrow -5^{+} } f(x) = +\infty\) The function \(f(x)=\frac{x-5}{x^{2}-25}\) has a vertical asymptote at \(x=-5\) since the limits approach infinity on both sides, confirming the singularity.

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Key Concepts

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

Limits
In calculus, a **limit** is a fundamental concept that describes the behavior of a function as its input approaches a particular value. To understand this idea, imagine approaching a destination point on a road. As you get nearer, the limit tells you what the road does鈥攚hether it continues smoothly, ends, or takes a sharp turn.

With the function \(f(x)=\frac{x-5}{x^2-25}\), we analyzed the behavior as \(x\) approached certain values to find the limits. For instance:
  • As \(x\) moves towards 5, the function remains smooth, simplifying to \(\frac{1}{10}\), indicating continuity rather than a vertical asymptote鈥攐nly a removable discontinuity or a 'hole'.
  • In contrast, approaching \(-5\) from the left or right, the function behaves differently, with limits heading towards \(-\infty\) or \(+\infty\) respectively.
Tracking these limits is essential for understanding where a function spikes up infinitely, which characterizes vertical asymptotes.
Rational Functions
**Rational functions** are expressions of the form \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials. These functions can exhibit a variety of behaviors, including continuity, points of discontinuity, and asymptotes.

In the example provided, \(f(x) = \frac{x-5}{x^2-25}\) is broken down into simpler factors by recognizing the denominator as a difference of squares: \((x-5)(x+5)\).
  • After canceling common factors in the numerator and the denominator, the original representation highlights that \(x=5\) leads to a removable discontinuity (a hole), not an asymptote.
  • Recognizing this factorization enables us to understand the function better by identifying potential points for vertical asymptotes, such as \(x=-5\) where the denominator equals zero.
Grasping these characteristics helps in predicting and detailing the behavior of rational functions under various conditions.
Asymptotic Behavior
**Asymptotic behavior** in mathematics describes how functions behave as they move towards infinitely large or small values. This notion is particularly important in understanding vertical asymptotes, where a function tends to increase or decrease indefinitely as it approaches a certain value of \(x\).

With the given function \(f(x)=\frac{x-5}{x^2-25}\), studying its asymptotic behavior involves recognizing where the function dramatically spikes towards positive or negative infinity:
  • For \(x\rightarrow -5\), the reduced function \(f(x) = \frac{1}{x+5}\) heads rapidly towards infinity, highlighting a vertical asymptote at \(x=-5\).
  • As \(x\) approaches this asymptote from the left, the function dips accordingly to \(-\infty\), while from the right, it rises to \(+\infty\).
Understanding this asymptotic nature is crucial in mathematics, providing insights into the limits and boundaries of a function, and predicting its behavior near specific values.

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Most popular questions from this chapter

a. Evaluate \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x),\) and then identify any horizontal asymptotes. b. Find the vertical asymptotes. For each vertical asymptote \(x=a\), evaluate \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\). $$f(x)=16 x^{2}\left(4 x^{2}-\sqrt{16 x^{4}+1}\right)$$

The hyperbolic cosine function, denoted cosh \(x\), is used to model the shape of a hanging cable (a telephone wire, for example). It is defined as \(\cosh x=\frac{e^{x}+e^{-x}}{2}\). a. Determine its end behavior by evaluating \(\lim \cosh x\) and \(\lim _{x \rightarrow-\infty} \cosh x\). b. Evaluate cosh 0. Use symmetry and part (a) to sketch a plausible graph for \(y=\cosh x\).

If a function \(f\) represents a system that varies in time, the existence of \(\lim _{t \rightarrow \infty} f(t)\) means that the system reaches a steady state (or equilibrium). For the following systems, determine if a steady state exists and give the steady-state value. The population of a culture of tumor cells is given by \(p(t)=\frac{3500 t}{t+1}\).

Find polynomials \(p\) and \(q\) such that \(p / q\) is undefined at 1 and \(2,\) but \(p / q\) has a vertical asymptote only at \(2 .\) Sketch a graph of your function.

The hyperbolic sine function is defined as \(\sinh x=\frac{e^{x}-e^{-x}}{2}\). a. Determine its end behavior by evaluating \(\lim \sinh x\) and \(\lim _{x \rightarrow-\infty} \sinh x\). b. Evaluate sinh 0. Use symmetry and part (a) to sketch a plausible graph for \(y=\sinh x\).
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