Question

Use l'Hôpital's Rule to evaluate the following limits. $$\lim _{x \rightarrow 0^{+}}(\tanh x)^{x}$$

Short answer

Answer: The limit of \((\tanh x)^x\) as \(x\) approaches \(0\) from the right is 1.

Step-by-step solution

Rewrite the expression into a suitable form for l'Hôpital's Rule

Let's rewrite the given expression using the property: $$a^b = e^{b \ln a}$$ So, $$\lim _{x \rightarrow 0^{+}}(\tanh x)^{x} = \lim _{x \rightarrow 0^{+}} e^{x \ln (\tanh x)}$$ Now, we have the expression inside the limit in the form of \(\frac{f(x)}{g(x)}\) since computing \(x \rightarrow 0^{+}\) is indeterminate form \(0{\cdot}\infty\) or the extended form \(\frac{\infty}{\infty}\) and we must derive the new function to compute the limit.

Differentiate both the numerator and the denominator

We need to compute the derivative for both the numerator and the denominator separately. Let's start with the numerator: $$f(x) = x \ln (\tanh x)$$ Apply the product rule to differentiate \(f(x)\): $$f'(x) = x'\ln(\tanh x)+x\ln'(\tanh x) = \ln(\tanh x)+x\frac{1}{\tanh x}\left(\frac{\operatorname{sech}^2 x}{\cosh x}\right)$$ Now, differentiate the denominator which is \(g(x)=x\): $$g'(x) = 1$$

Apply l'Hôpital's Rule to evaluate the limit for the new function

Now we apply l'Hôpital's rule: $$L = e^{\lim_{x \rightarrow 0^{+}}\frac{f'(x)}{g'(x)}}=e^{\lim_{x \rightarrow 0^{+}}\left[\ln(\tanh x)+x\frac{1}{\tanh x}\left(\frac{\operatorname{sech}^2 x}{\cosh x}\right)\right]}$$

Compute the limit

To compute the limit as x approaches 0, let's analyze each term: $$\lim_{x \rightarrow 0^{+}}\ln(\tanh x) = \ln(\tanh(0)) = \ln(0) = -\infty$$ Substitute this into the expression: $$L = e^{\lim_{x \rightarrow 0^{+}}\left[-\infty + x\frac{1}{\tanh x}\left(\frac{\operatorname{sech}^2 x}{\cosh x}\right)\right]}$$ Now, we need to analyze the second term: $$L = e^{\lim_{x \rightarrow 0^{+}} x\frac{1}{\tanh x}\left(\frac{\operatorname{sech}^2 x}{\cosh x}\right)}$$ Notice that as \(x \rightarrow 0^{+}\), the expression becomes \(\infty{\cdot}0{\cdot}\frac{1}{1}=0\). Thus, we get: $$L = e^0 = 1$$ So, after applying l'Hôpital's rule and evaluating the limit, we found that: $$\lim _{x \rightarrow 0^{+}}(\tanh x)^{x} = 1$$

Key concepts

Indeterminate Forms

Indeterminate forms often occur when evaluating limits and result in expressions where more information is needed to determine the exact value. An indeterminate form implies that at a first glance, the limit expression might suggest multiple possibilities for a limit value, such as 0/0 or ∞/∞. These types of expressions indicate that simple substitution does not suffice, so further mathematical methods or techniques must be applied to resolve them.

In the original exercise, we face an indeterminate form while evaluating \[ \lim_{x \to 0^{+}}(\tanh x)^{x} \] as x approaches 0. Its expression initially appears as 0·∞ or more traditionally understood through the transformed form \( e^{\lim_{x \to 0^{+}} x \ln(\tanh x)} \), leading us to an expression like ∞/∞. Instead of direct computation, such indeterminate forms require the application of a specific strategy, such as L'Hôpital's Rule, to find the limit.

L'Hôpital's Rule becomes an essential tool, helping us manage these indeterminate forms by switching our focus to the derivatives of the involved functions, ensuring that we arrive at a solution.

Exponential Function

The exponential function plays a pivotal role in our exercise, allowing us to transform the expression into a manageable form for limit calculations. The fundamental rule here is that converting a power expression \( a^b \) into \( e^{b \ln a} \) helps accommodate the application of different calculus techniques, such as L'Hôpital's Rule. This transformation leverages the exponential function where, essentially, any number raised to a power (the base 'e') is key in addressing complex limits.

Using the exponential function in the limit expression, \[\lim _{x \rightarrow 0^{+}} e^{x \ln (\tanh x)} \], allows us to analyze the base \( \tanh x \) and its impact as \( x \) approaches zero more systematically. Rewriting with the exponential function simplifies what would be an unwieldy limit, making it possible to utilize established calculus techniques to find a solution.

This strategy not only aids in applying L'Hôpital's Rule but also supports clearer differentiation of expressions otherwise represented through powers, lending a standardized approach to solving such problems.

Hyperbolic Functions

Hyperbolic functions, such as \( \tanh x \), are similar to trigonometric functions but relate to hyperbolas rather than circles. The hyperbolic tangent function, \( \tanh x = \frac{\sinh x}{\cosh x} \), is essential in our original problem as it forms the basis element in examining the limit expression.

Analyzing the behavior of \( \tanh x \) as \( x \to 0^{+} \) helps us understand the progression toward zero, critically affecting the expression’s limit. Understanding hyperbolic functions' properties, such as their limits and derivatives, lays the framework for evaluating more complex forms, allowing us to deconstruct seemingly intricate expressions into manageable calculations.

The role of hyperbolic functions becomes particularly noticeable when differentiating \( x \ln(\tanh x) \). The derivative involves both \( \operatorname{sech}^2 x \) and \( \cosh x \), again showing how these functions facilitate the handling of involved calculus operations within L'Hôpital's Rule's context. Hyperbolic functions, thus, serve as both foundational elements of the expression and central components enabling detailed calculus explorations.