Question

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

Short answer

Answer: The value of the limit is 0.

Step-by-step solution

Rewrite the expression

In order to apply l'Hôpital's Rule to the given limit, we first need to rewrite it as a quotient of two functions. We can do this using the natural logarithm and exponentiating the expression: \(\lim _{x \rightarrow 0^{+}}(\tanh x)^{x} = \lim _{x \rightarrow 0^{+}}e^{\ln\left((\tanh x)^{x}\right)}\) Now, we can use the property of logarithms that states \(\ln(a^b) = b\cdot \ln(a)\): \(\lim _{x \rightarrow 0^{+}}e^{\ln\left((\tanh x)^{x}\right)} = \lim _{x \rightarrow 0^{+}}e^{x\cdot\ln(\tanh x)}\) So, we need to find the limit of \(x\cdot\ln(\tanh x)\) as \(x \rightarrow 0^{+}\), and then exponentiate the result.

Separate the limit into a quotient

To apply l'Hôpital's Rule, we need to rewrite the expression as a fraction. To do that, we use the fact that \(a\cdot \ln(b) = \frac{\ln(b^a)}{1/a}\). Therefore: \(\lim _{x \rightarrow 0^{+}}x\cdot\ln(\tanh x) = \lim _{x \rightarrow 0^{+}}\frac{\ln(\tanh x)}{\frac{1}{x}}\)

Apply l'Hôpital's Rule

Now, we need to compute the derivatives of the numerator and the denominator separately for the newly-created quotient: Numerator derivative: \(\frac{d}{dx}(\ln(\tanh x)) = \frac{ sech^2(x)}{\tanh(x)}\) Denominator derivative: \(\frac{d}{dx}(\frac{1}{x})=-\frac{1}{x^2}\) Now we create a new limit expression with these derivatives in the form \(\frac{u}{v}\) and calculate the limit: \(\lim _{x \rightarrow 0^{+}}\frac{\frac{ sech^2(x)}{\tanh(x)}}{-\frac{1}{x^2}}\)

Simplify and find the limit

First, we simplify the expression: \(\lim _{x \rightarrow 0^{+}}\frac{\frac{ sech^2(x)}{\tanh(x)}}{-\frac{1}{x^2}} = \lim _{x \rightarrow 0^{+}}-\frac{ sech^2(x)\cdot x^2}{\tanh(x)}\) Now, as \(x \rightarrow 0^{+}\), we have \(\tanh(x) \rightarrow 0^{+}\) and \(sech^2(x)\cdot x^2 \rightarrow 0\). However, we can simplify further using the identity \(sech^2(x)=1-\tanh^2(x)\): \(\lim_{x \rightarrow 0^{+}}-\frac{(1-\tanh^2(x))x^2}{\tanh(x)}\) Now we can take the limit of each term separately: \(\lim_{x \rightarrow 0^{+}}-\frac{(1-\tanh^2(x))x^2}{\tanh(x)} = \lim_{x \rightarrow 0^{+}}-\frac{x^2}{\tanh(x)} = -\infty\)

Find the original limit

Recall that we want to find the limit of \(e^{x\cdot\ln(\tanh x)}\). Since the limit of \(x\cdot\ln(\tanh x)\) approaches \(-\infty\), we have: \(\lim_{x \rightarrow 0^{+}}(\tanh x)^{x} = \lim_{x \rightarrow 0^{+}}e^{\lim_{x \rightarrow 0^{+}}x\cdot\ln(\tanh x)} = \lim_{x \rightarrow 0^{+}}e^{-\infty} = 0\) Therefore, the limit is equal to 0.

Key concepts

Limits of Functions

Limits help us understand the behavior of a function as the input approaches a certain value. This concept is crucial when dealing with expressions that become indeterminate, meaning their values are not defined, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
When we deal with the hyperbolic tangent function, \(\tanh(x)\), we find its limit by examining the behavior of the expression as \(x\) approaches zero from the positive side, denoted by \(x \rightarrow 0^+\).
In this exercise, the limit of \((\tanh x)^x\) as \(x\) approaches zero is investigated. We convert the original problem using natural logarithms and exponentials to facilitate the application of l'Hôpital's Rule, a handy tool for tackling these indeterminate forms.

Natural Logarithms

A natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\), where \(e\) is an irrational number approximately equal to 2.71828. Natural logarithms are integral in mathematical transformation techniques that simplify complex expressions.
In the context of this exercise, we start by rewriting \((\tanh x)^x\) using the property of logarithms: \(\ln(a^b) = b\cdot \ln(a)\). This makes the expression easier to work with:

• \(\ln((\tanh x)^x) = x \cdot \ln(\tanh x)\). We will focus on finding the limit of this expression.

Transformations with log and exp functions often uncover simpler forms, allowing limits to be more easily evaluated.

Derivatives

Derivatives tell us about how functions change, giving the rate of change or slope at any point. They're essential for applying l'Hôpital's Rule, which allows us to evaluate indeterminate limits by taking the derivatives of the numerator and the denominator.
In this step-by-step solution, we first express the limit as a fraction and differentiate both components:

• The numerator: \(\ln(\tanh x)\) differentiates to \(\frac{sech^2(x)}{\tanh(x)}\).
• The denominator: \(\frac{1}{x}\) differentiates to \(-\frac{1}{x^2}\).

Using these derivatives, we create a new limit expression and simplify, enabling us to find the behavior of the function as \(x\) approaches zero. Thus, we acquire the final answer, which is essential for understanding how \((\tanh x)^x\) approaches zero as \(x\) decreases to zero.