Question

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

Short answer

Answer: The value of the limit is \(\frac{2}{\pi}\).

Step-by-step solution

Determine the indeterminate form of the limit

To apply l'Hôpital's Rule, we need to evaluate the given limit and check if it is in indeterminate form, i.e., \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\): $$\lim _{x \rightarrow 0} \frac{\tanh ^{-1} x}{\tan (\pi x / 2)}$$ As \(x \rightarrow 0\), both the numerator and the denominator approach 0. Therefore, the limit is in the indeterminate form \(\frac{0}{0}\).

Find the derivatives of the numerator and the denominator

We will now find the derivatives of the given functions with respect to x: Numerator: \(\frac{d}{dx}(\tanh^{-1}x)\) Denominator: \(\frac{d}{dx}(\tan(\pi x / 2))\) Using chain rule, we find the derivatives as follows: $$\frac{d}{dx}(\tanh^{-1}x) = \frac{1}{1-x^2}$$ $$\frac{d}{dx}(\tan(\pi x / 2)) = \frac{\pi}{2}\sec^2(\pi x/2)$$

Apply l'Hôpital's Rule

Now, we can apply l'Hôpital's Rule by taking the new limit of the ratio of derivatives: $$\lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\tanh^{-1}x)}{\frac{d}{dx}(\tan(\pi x / 2))}=\lim _{x \rightarrow 0} \frac{\frac{1}{1-x^2}}{\frac{\pi}{2}\sec^2(\pi x/2)}$$

Evaluate the new limit

We will now evaluate this new limit as \(x \rightarrow 0\): $$\lim _{x \rightarrow 0} \frac{\frac{1}{1-x^2}}{\frac{\pi}{2}\sec^2(\pi x/2)} = \frac{\frac{1}{1-0^2}}{\frac{\pi}{2}\sec^2(\pi \cdot 0/2)} = \frac{1}{\frac{\pi}{2}} = \frac{2}{\pi}$$

Final result

Therefore, the limit is equal to: $$\lim _{x \rightarrow 0} \frac{\tanh ^{-1} x}{\tan (\pi x / 2)} = \frac{2}{\pi}$$

Key concepts

Indeterminate Forms

Indeterminate forms occur when attempting to evaluate limits that have no clear resolution using standard limit laws. They often arise in mathematical operations involving division, multiplication, or powers, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms can be quite challenging because they do not immediately indicate any particular limit value through simple substitution.
In the problem provided, the expression \(\lim _{x \rightarrow 0} \frac{\tanh ^{-1} x}{\tan (\pi x / 2)}\) becomes \(\frac{0}{0}\) as \(x\) approaches zero. This is a classic indeterminate form.

• It is essential to recognize indeterminate forms before applying techniques like l'Hôpital's Rule, which helps resolve these ambiguous cases.
• l'Hôpital's Rule states that if a limit is in the form of \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), taking the derivative of the numerator and the denominator can provide a resolvable limit.

Chain Rule

The chain rule is a fundamental technique in calculus for finding derivatives of composite functions. When a function is composed of two or more functions, the chain rule provides a way to differentiate it by considering the derivative of the outer function and multiplying it by the derivative of the inner function.
To apply the chain rule, remember the format: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).
In the given exercise:

• The function \(\tanh^{-1} x\) is a simple inverse function, and its derivative is \(\frac{1}{1-x^2}\).

For \(\tan (\pi x / 2)\), we utilize the chain rule:

• The outer function is \(\tan(x)\) with derivative \(\sec^2(x)\).
• The inner function is \(\pi x / 2\) with derivative \(\pi / 2\).

This gives the derivative \(\frac{d}{dx}(\tan(\pi x / 2)) = \frac{\pi}{2} \sec^2(\pi x/2)\).
Using the chain rule effectively requires knowledge of basic derivative rules and being able to identify the nested functions involved.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, like \(\tanh^{-1}x\), are used to provide a "reverse" function for hyperbolic trigonometric functions. They serve similar purposes as inverse trigonometric functions but are used for hyperbolic functions, which are based on exponential functions rather than circular functions.
These functions have specific derivative formulas that make evaluating limits involving them possible. For example, the derivative of \(\tanh^{-1}x\) is \(\frac{1}{1-x^2}\).

• Inverse hyperbolic functions are often seen in calculus problems due to their unique properties and occurrences in growth and decay models.
• Understanding their derivatives is crucial because they often pop up when evaluating limits or integrals.

In the exercise, \(\tanh^{-1}x\) contributed to the indeterminate form from which l'Hôpital's Rule was applied, requiring its derivative for successful limit evaluation.