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Chapter 4: Problem 49

Evaluate the following limits. $$\lim _{x \rightarrow \pi / 2^{-}}\left(\frac{\pi}{2}-x\right) \sec x$$

Short Answer

Expert verified
Answer: The limit is 1.

Step by step solution

01

Identify the trigonometric function

We need to examine the limit of the given function that contains the secant function. Recall that sec(x) = 1/cos(x). Step 2: Rewrite sec(x) in terms of cos(x)
02

Rewrite sec(x)

Write the given function in terms of cosine by replacing sec(x) with 1/cos(x). $$\lim _{x \rightarrow \pi / 2^{-}}\left(\frac{\pi}{2}-x\right) \frac{1}{\cos x}$$ Step 3: Analyze the behavior of the numerator and the denominator
03

Analyze the behavior

As x gets closer to 蟺/2 from the left side, the numerator (蟺/2 - x) approaches 0. The denominator (cos(x)) also approaches 0, because cos(蟺/2) = 0. Step 4: Use L'H么pital's Rule
04

Apply L'H么pital's Rule

Since we have the indeterminate form 0/0, we can apply L'H么pital's Rule, which states that: $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ If both \(\lim_{x \to a} f'(x)\) and \(\lim_{x \to a} g'(x)\) exist. Step 5: Differentiate the numerator and denominator
05

Differentiate

Differentiate the numerator and denominator with respect to x. $$f(x) = \frac{\pi}{2}-x \Rightarrow f'(x) = -1$$ $$g(x) = \cos{x} \Rightarrow g'(x) = -\sin{x}$$ Step 6: Evaluate the new limit
06

Evaluate the limit

Plug the derivatives back into the limit and evaluate. $$\lim _{x \rightarrow \pi / 2^{-}}\frac{-1}{-\sin x}$$ As x approaches 蟺/2 from the left side, sin(x) approaches 1, and the limit becomes: $$\lim _{x \rightarrow \pi / 2^{-}} \frac{-1}{-1} = 1$$ The evaluated limit is 1.

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