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Chapter 2: Problem 35

Are the functions continuous at the point being approached? \(\lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right)\) 3

Short Answer

Expert verified
Yes, the function is continuous at \( t = 0 \).

Step by step solution

01

Understand the Function

We need to check if the function \( f(t) = \cos\left(\frac{\pi}{\sqrt{19 - 3 \sec 2t}}\right) \) is continuous at \( t = 0 \). For this, we need to evaluate the behavior of \( \sec 2t \) as \( t \to 0 \).
02

Evaluate \( \sec 2t \) as \( t \to 0 \)

The function \( \sec 2t = \frac{1}{\cos 2t} \) will converge to \( \sec 0 = 1 \) since \( \cos 0 = 1 \). Substitute this into the original function.
03

Substitute into the function

Plugging \( \sec 0 = 1 \) into the function, we get \( \frac{\pi}{\sqrt{19 - 3\cdot 1}} \). Simplifying, this becomes \( \frac{\pi}{\sqrt{16}} = \frac{\pi}{4} \).
04

Evaluate the Cosine Function

Now substitute back into \( \cos\left(\frac{\pi}{4}\right) \). From trigonometric identities, \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).
05

Check and Conclude Continuity

The limit as \( t \to 0 \) exists and evaluates to a finite value \( \frac{\sqrt{2}}{2} \). Therefore, the function is continuous at \( t = 0 \).

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Key Concepts

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

Limits
In calculus, limits help us describe the behavior of a function as it approaches a certain point. Understanding limits is essential because they form the foundation for defining many other important concepts in calculus, such as derivatives and continuity.
  • A limit looks at what happens to function values as inputs get close to a specific point without necessarily touching the point.
  • For example, in our exercise, we're interested in what happens to the function \( f(t) = \cos\left(\frac{\pi}{\sqrt{19 - 3 \sec 2t}}\right) \) as \( t \to 0 \).
We determined that as \( t \to 0 \), \( \sec 2t \to 1 \) because \( \cos 2t \to 1 \). The ultimate goal is to simplify the expression and check if the function behaves nicely and doesn't break into pieces near the point we care about (\( t = 0 \) here).
Trigonometric Functions
Trigonometric functions like cosine, sine, and tangent are critical in calculus because they describe the geometry of waves and other cyclical patterns.
  • The cosine function, denoted as \( \cos \), outputs values ranging from -1 to 1 depending on the angle you input.
  • In our situation, \( \cos\left(\frac{\pi}{4}\right) \) returns \( \frac{\sqrt{2}}{2} \), which is a well-known trigonometric value often used in unit circle calculations.
Trigonometric identities, such as \( \cos(0) = 1 \), are pivotal in simplifying expressions and finding limits. Understanding these basic values and identities is key to evaluating limits involving trigonometric functions quickly and correctly.
Evaluating Limits
To evaluate limits, especially when involving trigonometric or other complex functions, a methodical approach is vital. Here's how we approached evaluating the limit in our example:
  • First, understand the function and identify tricky parts, like \( \sec 2t \).
  • Simplify by finding the behavior of components as the input approaches the point of interest, here \( t = 0 \).
  • Substitute these simplified behaviors back into the main function to resolve any indeterminate forms.
Evaluating \( \lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2t}}\right) \), we examined how \( \sec 2t \) simplified to 1, then calculated further to find the limit of the whole function. Each simplification step brings us closer to determining if the function reaches a steady, continuous outcome. This process shows whether a function is continuous at a point, and if there are no jumps or holes in the function's value at that location.

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Most popular questions from this chapter

You will find a graphing calculator useful for Exercise. Let \(f(x)=\left(3^{x}-1\right) / x.\) a. Make tables of values of \(f\) at values of \(x\) that approach \(c=0\) from above and below. Does \(f\) appear to have a limit as \(x \rightarrow 0 ?\) If so, what is it? If not, why not? b. Support your conclusions in part (a) by graphing \(f\) near \(c=0.\)

Graph the rational functions .Include the graphs and equations of the asymptotes. $$y=\frac{x^{3}+1}{x^{2}}$$

a. Graph \(g(x)=x \sin (1 / x)\) to estimate \(\lim _{x \rightarrow 0} g(x),\) zooming in on the origin as necessary. b. Confirm your estimate in part (a) with a proof.

Gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon \geq 0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. $$f(x)=m x, \quad m>0, \quad L=3 m, \quad c=3, \quad \epsilon=c>0$$

If \(2-x^{2} \leq g(x) \leq 2 \cos x\) for all \(x,\) find \(\lim _{x \rightarrow 0} g(x).\)
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