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Chapter 2: Problem 51

Asymptotes Use analytical methods and/or a graphing utility to identify the vertical asymptotes (if any) of the following functions. $$f(x)=\frac{1}{\sqrt{x} \sec x}$$

Short Answer

Expert verified
Question: Determine the vertical asymptote(s) of the function \(f(x) = \frac{1}{\sqrt{x} \sec x}\). Answer: The function has one vertical asymptote at x = 0.

Step by step solution

01

Identify the denominator of the function

The given function is: $$f(x)=\frac{1}{\sqrt{x} \sec x}$$ The denominator of this function is \(\sqrt{x} \sec x\).
02

Determine the values of x that make the denominator equal to zero

In order to find the vertical asymptotes, we set the denominator equal to zero and solve for x: $$\sqrt{x} \sec x = 0$$ We need to analyze both the square root and the secant function separately in order to determine where the denominator is equal to zero.
03

Analyze the square root function

First, let's find the values of x that make the square root function equal to zero: $$\sqrt{x} = 0$$ The only value of x that satisfies this equation is x = 0.
04

Analyze the secant function

Now let's analyze the secant function. The secant function is the reciprocal of the cosine function: $$\sec x = \frac{1}{\cos x}$$ For \(\sec x\) to equal zero, the cosine function should be undefined. However, \(\cos x\) is always defined for real values of x. So there are no values of x that make \(\sec x = 0\).
05

Determine the vertical asymptotes

From steps 3 and 4, we found that the only value of x that makes the denominator zero is x = 0. So, we have a vertical asymptote at x = 0 if the numerator is not zero, which is true in this case. Therefore, the function has one vertical asymptote at x = 0.

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Most popular questions from this chapter

Evaluate the following limits. $$\lim _{x \rightarrow 1^{-}} \frac{x}{\ln x}$$

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 say that \(\lim _{x \rightarrow a} f(x) \neq L\) 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 \(\varepsilon=\frac{1}{2}\).)

Determine the end behavior of the following transcendental functions by evaluating appropriate limits. Then provide a simple sketch of the associated graph, showing asymptotes if they exist. $$f(x)=|\ln x|$$

Evaluate the following limits. $$\lim _{x \rightarrow 0^{+}} \frac{x}{\ln x}$$

Let $$g(x)=\left\\{\begin{array}{ll}x^{2}+x & \text { if } x<1 \\\a & \text { if } x=1 \\\3 x+5 & \text { if } x>1.\end{array}\right.$$ a. Determine the value of \(a\) for which \(g\) is continuous from the left at 1. b. Determine the value of \(a\) for which \(g\) is continuous from the right at 1. c. Is there a value of \(a\) for which \(g\) is continuous at \(1 ?\) Explain.
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