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Find the limits in Exercises \(47-54\) $$ \lim _{\theta \rightarrow \pi / 4} \frac{\tan \theta-1}{\theta-\frac{\pi}{4}} $$

Short Answer

Expert verified
The limit is 2.

Step by step solution

01

Identify Form of the Limit

The given expression is a limit problem of the form \( \lim _{\theta \rightarrow a} \frac{f(\theta)}{g(\theta)} \). Here, \( f(\theta) = \tan \theta - 1 \) and \( g(\theta) = \theta - \frac{\pi}{4} \) as \( \theta \rightarrow \frac{\pi}{4} \). We need to determine if this limit results in an indeterminate form like \( \frac{0}{0} \).
02

Check for Indeterminate Form

Substitute \( \theta = \frac{\pi}{4} \) into \( f(\theta) \) and \( g(\theta) \):- \( f(\frac{\pi}{4}) = \tan \frac{\pi}{4} - 1 = 1 - 1 = 0 \)- \( g(\frac{\pi}{4}) = \frac{\pi}{4} - \frac{\pi}{4} = 0 \)Both \( f(\theta) \) and \( g(\theta) \) are zero at \( \theta = \frac{\pi}{4} \), giving an indeterminate form \( \frac{0}{0} \). Therefore, L'Hospital's Rule can be applied.
03

Apply L'Hospital's Rule

L'Hospital's Rule states that if a limit is in the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), the limit of \( \frac{f(\theta)}{g(\theta)} \) as \( \theta \to a \) can be found by evaluating \( \lim _{\theta \rightarrow a} \frac{f'(\theta)}{g'(\theta)} \).
04

Derivatives of the Functions

Find the derivatives of \( f(\theta) \) and \( g(\theta) \):- \( f'(\theta) = (\tan \theta - 1)' = \sec^2 \theta \)- \( g'(\theta) = (\theta - \frac{\pi}{4})' = 1 \)
05

Calculate the Limit Using Derivatives

Substitute the derivatives back into the limit:\[ \lim _{\theta \rightarrow \pi/4} \frac{f'(\theta)}{g'(\theta)} = \lim _{\theta \rightarrow \pi/4} \frac{\sec^2 \theta}{1} = \lim _{\theta \rightarrow \pi/4} \sec^2 \theta\]At \( \theta = \frac{\pi}{4} \), \( \sec \frac{\pi}{4} = \sqrt{2} \). Therefore, \( \sec^2 \frac{\pi}{4} = 2 \).
06

Conclusion

The limit simplifies to 2, so:\[ \lim _{\theta \rightarrow \pi/4} \frac{\tan \theta - 1}{\theta - \frac{\pi}{4}} = 2 \]

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Key Concepts

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

Indeterminate Forms
In calculus, indeterminate forms are expressions that do not have a clear or definite result when evaluating limits. The most common indeterminate forms are \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), \( \infty - \infty \), and others that don’t provide a straightforward conclusion.
Understanding these forms is crucial because they often indicate that the limit cannot be determined through direct substitution. Instead, special techniques like algebraic manipulation or calculus-based methods, such as L'Hospital’s Rule, are required.
  • In our example, the limit expression \( \lim _{\theta \rightarrow \pi / 4} \frac{\tan \theta-1}{\theta-\frac{\pi}{4}} \) results in the \( \frac{0}{0} \) indeterminate form when directly substituting \( \theta = \frac{\pi}{4} \).
  • This discovery allows us to employ techniques such as L’Hospital’s Rule to evaluate the limit properly.
Limit Evaluation
Evaluating limits involves finding the value that a function approaches as the input approaches a given value. Many times, especially in calculus, limits are used to understand the behavior of functions that are not easily expressed at specific points.
Traditional limit evaluation through substitution might fail when encountering indeterminate forms. This is where calculus techniques like L'Hospital’s Rule become invaluable.

L'Hospital’s Rule can transform such tricky problems into more manageable ones by using derivatives. For example:
  • In the expression \( \lim _{\theta \rightarrow \pi / 4} \frac{\tan \theta-1}{\theta-\frac{\pi}{4}} \), direct substitution gives 0 in both the numerator and the denominator.
  • Recognizing this indeterminate form allows us to apply L'Hospital’s Rule.
  • By differentiating both the numerator and the denominator, the limit becomes \( \lim _{\theta \rightarrow \pi/4} \sec^2 \theta = 2 \).
  • Thus, after evaluating the derivatives, the limit simplifies to 2.
Derivatives
Derivatives are fundamental in calculus, representing the rate of change of a function with respect to one of its variables. They form the backbone of methods such as L'Hospital's Rule, which is used to solve limit problems involving indeterminate forms.

Understanding how to take derivatives is crucial when applying L'Hospital’s Rule. This technique requires differentiating both the numerator and the denominator of a fraction to evaluate a limit.
  • For our problem, the given functions are \( f(\theta) = \tan \theta - 1 \) and \( g(\theta) = \theta - \frac{\pi}{4} \).
  • The derivatives are \( f'(\theta) = \sec^2 \theta \) and \( g'(\theta) = 1 \).
  • These derivatives allow us to transform the original indeterminate form \( \frac{0}{0} \) into a simple evaluation of \( \sec^2 \theta \) as \( \theta \to \frac{\pi}{4} \).
  • Thus, knowing derivatives enables the effective use of L'Hospital's Rule.
By mastering derivatives, we better understand both the behavior of functions and the applicable limits, making problems that initially seem complex much simpler to handle.

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Most popular questions from this chapter

Diagonals If \(x, y,\) and \(z\) are lengths of the edges of a rectangular box, the common length of the box's diagonals is \(s=\) \(\sqrt{x^{2}+y^{2}+z^{2}}\) a. Assuming that \(x, y,\) and \(z\) are differentiable functions of \(t\) how is \(d s / d t\) related to \(d x / d t, d y / d t,\) and \(d z / d t\) ? b. How is \(d s / d t\) related to \(d y / d t\) and \(d z / d t\) if \(x\) is constant? c. How are \(d x / d t, d y / d t,\) and \(d z / d t\) related if \(s\) is constant?

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Area Suppose that the radius \(r\) and area \(A=\pi r^{2}\) of a circle are differentiable functions of \(t\) . Write an equation that relates \(d A / d t\) to \(d r / d t .\)

In Exercises \(57-60,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2 $$

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