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Chapter 3: Problem 52

Evaluate the following limits or state that they do not exist. $$\lim _{x \rightarrow \pi / 2} \frac{\cos x}{x-(\pi / 2)}$$

Short Answer

Expert verified
Question: Determine the limit of the function $$f(x)=\frac{\cos x}{x-(\pi / 2)}$$ as $$x\rightarrow \pi/2$$ using L'H么pital's Rule. Answer: The limit of the given function as $$x\rightarrow \pi/2$$ is -1.

Step by step solution

01

Identify the limit and check if it is indeterminate

Evaluate the function $$f(x)=\frac{\cos x}{x-(\pi / 2)}$$ as $$x\rightarrow \pi/2$$. $$\lim _{x \rightarrow \pi / 2} \frac{\cos x}{x-(\pi / 2)}$$ As $$x\rightarrow \pi/2$$, $$\cos x\rightarrow 0$$ and $$x-(\pi/2)\rightarrow 0$$. This limit is in the form of $$\frac{0}{0}$$, which is an indeterminate form.
02

Apply L'H么pital's Rule

Since we have an indeterminate form of $$\frac{0}{0}$$, apply L'H么pital's Rule. L'H么pital's Rule states: If $$\lim_{x\to a} \frac{f(x)}{g(x)}$$ is of the form $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$, and both $$f'(x)$$ and $$g'(x)$$ exist and are continuous around a, then $$\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}$$. Find the derivative of the numerator and denominator with respect to x: $$f'(x)=-\sin(x)$$ $$g'(x)=1$$
03

Evaluate the new limit

Replace the original function with the derivatives and evaluate the limit: $$\lim _{x \rightarrow \pi / 2} \frac{-\sin x}{1}$$ As $$x\rightarrow \pi/2$$, $$-\sin x\rightarrow -1$$ and the limit becomes: $$\lim _{x \rightarrow \pi / 2} \frac{-\sin x}{1} = -1$$ The limit of the given function is -1: $$\lim _{x \rightarrow \pi / 2} \frac{\cos x}{x-(\pi / 2)} = -1$$

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