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Chapter 6: Problem 86

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

Short Answer

Expert verified
Given function: $$\frac{\tanh^{-1}(x)}{\tan(\pi x/2)}$$ Answer: The limit of the given function as x approaches 1 from the left is 0.

Step by step solution

01

Find the derivative of the numerator function

In this case, the numerator function is the inverse hyperbolic tangent function, \(\tanh^{-1}(x)\). To find its derivative, we need to use the formula: $$\frac{d}{dx}\tanh^{-1}(x) = \frac{1}{1 - x^2}$$
02

Find the derivative of the denominator function

The denominator function is \(\tan(\frac{\pi x}{2})\). To find its derivative, we use the chain rule with the derivative of the tangent function (involving a constant multiple of x): $$\frac{d}{dx}\tan(\frac{\pi x}{2}) = \frac{\pi}{2}\cos^{-1}(\frac{\pi x}{2})$$
03

Rewrite the function as the ratio of the derivatives

Using l'H么pital's Rule, we can rewrite the given function as the ratio of the derivatives: $$\frac{\tanh^{-1}(x)}{\tan(\pi x / 2)} \Rightarrow \frac{\frac{1}{1 - x^2}}{\frac{\pi}{2}\cdot\cos^{-1}(\frac{\pi x}{2})}$$
04

Simplify the function

The function can be further simplified by multiplying both the numerator and denominator by the reciprocal of the denominator. This results in the following simplified function: $$\frac{2}{\pi}\cdot\frac{\cos(\frac{\pi x}{2})}{1 - x^2}$$
05

Evaluate the limit

Now, we can find the limit as x approaches 1 from the left: $$\lim_{x \rightarrow 1^-} \frac{2 \cos(\pi x / 2)}{\pi (1 - x^2)}$$ As x approaches 1, the cosine function in the numerator approaches 0, and the denominator approaches 0 as well. However, since we have already applied l'H么pital's Rule, the limit of this simplified function exists and equals: $$\frac{0}{0} \Rightarrow \boxed{0}$$ Therefore, the limit of the given function as x approaches 1 from the left is 0.

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