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Chapter 4: Problem 53

Evaluate the following limits. $$\lim _{\theta \rightarrow \pi / 2^{-}}(\tan \theta-\sec \theta)$$

Short Answer

Expert verified
Answer: The limit is 0.

Step by step solution

01

Recall the definitions of tangent and secant

Tan and sec are trigonometric functions defined as follows: $$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$ $$\sec(\theta) = \frac{1}{\cos(\theta)}$$ Substitute these definitions into the given limit expression: $$\lim _{\theta \rightarrow \pi / 2^-}(\frac{\sin(\theta)}{\cos(\theta)}-\frac{1}{\cos(\theta)})$$
02

Rewrite with a common denominator

To be able to combine the terms in the expression, rewrite them with a common denominator. The common denominator is \(\cos(\theta)\). Rewrite the expressions as: $$\lim _{\theta \rightarrow \pi / 2^-}(\frac{\sin(\theta)-1}{\cos(\theta)})$$
03

Evaluate the limit

Now, substitute the limit value \(\theta = \pi/2^{-}\) into the expression: $$\frac{\sin(\pi/2 )-1}{\cos(\pi/2 )}$$ Evaluate the sine and cosine functions at \(\theta = \pi/2^-\): $$\sin\left(\frac{\pi}{2}\right) = 1$$ $$\cos\left(\frac{\pi}{2}\right) = 0$$ Substitute these values back into the expression: $$\frac{1-1}{0}$$ Which simplifies to: $$\frac{0}{0}$$
04

Use L'Hopital's Rule

Since the limit resulted in an indeterminate form \(\frac{0}{0}\), we can use L'Hopital's Rule, which states that if \(\lim_{\theta\to a} \frac{f(\theta)}{g(\theta)}=\frac{0}{0}\), then: $$\lim_{\theta\to a} \frac{f(\theta)}{g(\theta)}=\lim_{\theta\to a} \frac{f'(\theta)}{g'(\theta)}$$ Differentiate the numerator and denominator: $$f'(\theta) = -\cos(\theta)$$ $$g'(\theta) = \sin(\theta)$$ Apply L'Hopital's Rule: $$\lim_{\theta\to \pi/2^-} \frac{-\cos(\theta)}{\sin(\theta)}$$
05

Evaluate the limit with L'Hopital's Rule

Evaluate the limit after applying L'Hopital's rule: $$\frac{-\cos(\pi/2)}{\sin(\pi/2)}$$ Which gives: $$\frac{-0}{1}$$ Thus, the limit evaluates to 0: $$\lim _{\theta \rightarrow \pi / 2^{-}}(\tan \theta-\sec \theta) = 0$$

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