Problem 85 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: Verify the conditions for applying l'Hôpital's Rule

To apply l'Hôpital'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 ∞/∞. 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.

Step 3: 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: $$g'(x) = (\pi / 2) \cos^2(\pi x / 2)$$

Step 4: Take the limit as x approaches 0 of the ratio of the derivatives

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

Step 5: 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: $$\lim_{x \rightarrow 0} \frac{\tanh^{-1}x}{\tan(\pi x / 2)} = \frac{2}{\pi}$$