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Chapter 4: Problem 39

Evaluate the following limits. $$\lim _{x \rightarrow \pi / 2^{-}} \frac{\tan x}{3 /(2 x-\pi)}$$

Short Answer

Expert verified
Answer: The limit as x approaches 蟺/2 from the left side of the given function is 0.

Step by step solution

01

Rewrite the function using properties of trigonometry

We can rewrite the function in the form of sine and cosine, which can sometimes help make the limit easier to evaluate: $$f(x) = \frac{\sin x}{\cos x} \cdot \frac{2x-\pi}{3}$$
02

Apply L'H么pital's Rule

Since we have a fraction with the limit approaching 蟺/2 from the left side, we can apply L'H么pital's Rule if we have an indeterminate form of 0/0 or 鈭�/鈭�. L'H么pital's Rule states that if the limit of the derivative of the numerator divided by the derivative of the denominator exists, then the limit of the function exists and is equal to that value. Take the derivative of the numerator: $$\frac{d}{dx} (\sin x) = \cos x$$ Take the derivative of the denominator: $$\frac{d}{dx} \left(\frac{2x-\pi}{3}\right) = \frac{2}{3}$$ So, our new function is: $$g(x) = \frac{\cos x}{\frac{2}{3}}$$
03

Evaluate limit of g(x) as x approaches 蟺/2 from the left side

Now, we can evaluate the limit of the new function: $$\lim _{x \rightarrow \pi / 2^{-}} g(x) = \lim _{x \rightarrow \pi / 2^{-}} \frac{\cos x}{\frac{2}{3}}$$ As x approaches 蟺/2, cos(x) approaches 0. Therefore, the limit is: $$\lim _{x \rightarrow \pi / 2^{-}} g(x) = \frac{0}{\frac{2}{3}} = 0$$ Since the limit of the derivatives exists and is equal to 0, we can conclude that the limit of the original function also exists and is equal to 0: $$\lim _{x \rightarrow \pi / 2^{-}} \frac{\tan x}{3 /(2 x-\pi)} = 0$$

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