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Chapter 2: Problem 3

For what values of \(a\) does \(\lim r(x)=r(a)\) if \(r\) is a rational function?

Short Answer

Expert verified
Answer: The values of \(a\) for which the limit of a rational function exists and equals the function value at that point are all the values in the domain of the rational function, i.e., all the values of \(x\) for which the denominator of the rational function is nonzero.

Step by step solution

01

Identify the continuity of the rational function

A rational function is continuous at all points where its denominator is nonzero. So, we need to identify the points where \(Q(x) \neq 0\), i.e., the domain of \(r(x)\).
02

Evaluate the limit

Since rational functions are continuous on their domain, the limit exists and equals the function value at that point. In other words, \(\lim_{x \to a} r(x) = r(a)\) for all \(a\) in the domain of \(r(x)\).
03

Find the domain of the rational function

To find the domain of the rational function \(r(x) = \frac{P(x)}{Q(x)}\), we need to find the values of \(x\) for which \(Q(x) \neq 0\). This will depend on the specific polynomial functions \(P(x)\) and \(Q(x)\) in the given rational function.
04

Determine the values of \(a\)

The values of \(a\) for which \(\lim_{x \to a} r(x) = r(a)\) are all the values in the domain of the rational function \(r(x)\). This means, the answer to the problem will depend on the specific rational function and the values of \(x\) for which its denominator is nonzero. In summary, the values of \(a\) for which the limit of a rational function exists and equals the function value at that point are all the values in the domain of the rational function, i.e., all the values of \(x\) for which the denominator of the rational function is nonzero.

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Most popular questions from this chapter

See Exercises 95-96. a. Does the function \(f(x)=x \sin (1 / x)\) have a removable discontinuity at \(x=0 ?\) b. Does the function \(g(x)=\sin (1 / x)\) have a removable discontinuity at \(x=0 ?\)

$$\begin{aligned} &\text {a. Use a graphing utility to estimate } \lim _{x \rightarrow 0} \frac{\tan 2 x}{\sin x}, \lim _{x \rightarrow 0} \frac{\tan 3 x}{\sin x}, \text { and }\\\ &\lim _{x \rightarrow 0} \frac{\tan 4 x}{\sin x} \end{aligned}$$ b. Make a conjecture about the value of \(\lim _{x \rightarrow 0} \frac{\tan p x}{\sin x},\) for any real constant \(p\)

Use analytical methods to identify all the asymptotes of \(f(x)=\frac{\ln \left(9-x^{2}\right)}{2 e^{x}-e^{-x}} .\) Then confirm your results by locating the asymptotes with a graphing utility.

Use the following definition for the nonexistence of a limit. Assume \(f\) is defined for all values of \(x\) near a, except possibly at a. We write \(\lim _{x \rightarrow a} f(x) \neq L\) if for some \(\varepsilon>0\) there is no value of \(\delta>0\) satisfying the condition $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0<|x-a|<\delta.$$ Let $$f(x)=\left\\{\begin{array}{ll} 0 & \text { if } x \text { is rational } \\ 1 & \text { if } x \text { is irrational. } \end{array}\right.$$ Prove that \(\lim _{x \rightarrow a} f(x)\) does not exist for any value of \(a\). (Hint: Assume \(\lim _{x \rightarrow a} f(x)=L\) for some values of \(a\) and \(L\) and let \(\varepsilon=\frac{1}{2}\).)

It can be shown that \(1-\frac{x^{2}}{6} \leq \frac{\sin x}{x} \leq 1,\) for \(x\) near 0 a. Illustrate these inequalities with a graph. b. Use these inequalities to evaluate \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\).
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