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Chapter 6: Problem 85

Use l'H么pital's Rule to evaluate the following limits. $$\lim _{x \rightarrow 0} \frac{\tanh ^{-1} x}{\tan (\pi x / 2)}$$

Short Answer

Expert verified
Answer: The limit of the function as x approaches 0 is $$\frac{2}{\pi}$$.

Step by step solution

01

Identify the given function and the limit as x approaches 0

We are given the function: $$\frac{\tanh^{-1}x}{\tan(\pi x / 2)}$$ And we need to find the limit as x approaches 0.
02

Verify the conditions for applying l'H么pital's Rule

To apply l'H么pital's Rule, we need to ensure that the limit of the ratio of the functions is of indeterminate form, which is either 0/0 or 鈭�/鈭�. Let's check that this is the case: $$\lim_{x \rightarrow 0} \tanh^{-1}x = 0$$ $$\lim_{x \rightarrow 0} \tan(\pi x / 2) = 0$$ Thus, the given limit is in the indeterminate form of 0/0.
03

Find the derivative of the numerator and the denominator

Now, we will find the derivative of the numerator and the denominator with respect to x: The derivative of the numerator, \(f(x) = \tanh^{-1}x\), is: $$f'(x) = \frac{1}{1 - x^2}$$ The derivative of the denominator, \(g(x) = \tan(\pi x / 2)\), is: $$g'(x) = (\pi / 2) \cos^2(\pi x / 2)$$
04

Take the limit as x approaches 0 of the ratio of the derivatives

Applying l'H么pital's Rule, we now take the limit as x approaches 0 of the ratio of the derivatives: $$\lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 0} \frac{\frac{1}{1 - x^2}}{(\pi / 2) \cos^2(\pi x / 2)}$$
05

Simplify the expression and evaluate the limit as x approaches 0

Let's simplify the expression and evaluate the limit: $$\lim_{x \rightarrow 0} \frac{1}{1 - x^2} \cdot \frac{2}{\pi \cos^2(\pi x / 2)}$$ Now, as x approaches 0: $$\lim_{x \rightarrow 0} \frac{1}{1 - 0^2} \cdot \frac{2}{\pi \cos^2(\pi \cdot 0 / 2)} = 1 \cdot \frac{2}{\pi} = \frac{2}{\pi}$$ Therefore, the limit of the given function as x approaches 0 is: $$\lim_{x \rightarrow 0} \frac{\tanh^{-1}x}{\tan(\pi x / 2)} = \frac{2}{\pi}$$

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Most popular questions from this chapter

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