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Chapter 3: Problem 55

Evaluate the following limits or state that they do not exist. $$\lim _{x \rightarrow \pi / 4} 3 \csc 2 x \cot 2 x$$

Short Answer

Expert verified
Answer: The limit of the function as x approaches 蟺/4 is 0.

Step by step solution

01

Simplify the function

To start, rewrite the given function in terms of sine and cosine functions: $$3 \csc 2x \cot 2x = 3 \cdot \frac{1}{\sin(2x)} \cdot \frac{\cos(2x)}{\sin(2x)}$$ Now, we can simplify the function: $$3 \cdot \frac{\cos(2x)}{\sin^2(2x)}$$
02

Use double angle identity

Rewrite the function using double angle identities for sine and cosine: The double angle formulas are: $$\sin(2x) = 2 \sin(x) \cos(x)$$ $$\cos(2x) = \cos^2(x) - \sin^2(x)$$ Now, plug these identities into our function: $$3 \cdot \frac{\cos^2(x) - \sin^2(x)}{(2 \sin(x) \cos(x))^2}$$
03

Simplify the function further

We have: $$3 \cdot \frac{\cos^2(x) - \sin^2(x)}{4\sin^2(x)\cos^2(x)}$$ Divide the numerator and denominator by \(\cos^2(x)\): $$3 \cdot \frac{1 - \tan^2(x)}{4\sin^2(x)}$$
04

Evaluate the limit

We now want to find the limit as x approaches 蟺/4: $$\lim_{x\rightarrow \pi/4} 3 \cdot \frac{1 - \tan^2(x)}{4\sin^2(x)}$$ At x = 蟺/4: - \(\tan(\pi/4) = 1\) - \(\sin(\pi/4) = \frac{1}{\sqrt{2}}\) Plug these values into the function: $$3 \cdot \frac{1 - 1^2}{4 \cdot \left(\frac{1}{\sqrt{2}}\right)^2}$$ Simplify: $$3 \cdot \frac{0}{4 \cdot \frac{1}{2}}$$ Therefore, the limit of the function as x approaches 蟺/4 is: $$\lim_{x\rightarrow \pi/4} 3 \csc 2x \cot 2x = 0$$

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