Question

Use l'Hôpital's Rule to evaluate the following limits. \(\lim _{x \rightarrow \infty} \frac{1-\operatorname{coth} x}{1-\tanh x}\)

Short answer

Question: Evaluate the limit of the function \(\frac{1-\operatorname{coth} x}{1-\tanh x}\) as \(x\) approaches infinity using l'Hôpital's Rule. Answer: The limit of the given function as \(x\) approaches infinity is 1.

Step-by-step solution

Differentiate the numerator and denominator

Let's start by differentiating the numerator and denominator with respect to \(x\): The derivative of \(\operatorname{coth} x = \frac{d}{dx} \operatorname{coth} x = -\operatorname{csch}^2 x\) The derivative of \(\tanh x = \frac{d}{dx} \tanh x = \operatorname{sech}^2 x \) Now, we have the derivatives of coth x and tanh x: \(-\operatorname{csch}^2 x\) and \(\operatorname{sech}^2 x\) Let's differentiate the numerator and denominator: $$\frac{d}{dx} (1-\operatorname{coth} x) = -\operatorname{csch}^2 x$$ $$\frac{d}{dx} (1-\tanh x) = -\operatorname{sech}^2 x$$

Compute the limit of the derivatives

We now need to compute the limits of the derivatives as \(x\) approaches infinity: $$\lim _{x \rightarrow \infty} -\operatorname{csch}^2 x = 0$$ $$\lim _{x \rightarrow \infty} -\operatorname{sech}^2 x = 0$$

Apply l'Hôpital's Rule

Since the limits of both derivatives equal to zero, we can now apply l'Hôpital's Rule. The limit of our function is equal to the limit of the quotient of the derivatives of the numerator and denominator: $$\lim _{x \rightarrow \infty} \frac{1-\operatorname{coth} x}{1-\tanh x} = \lim _{x \rightarrow \infty} \frac{-\operatorname{csch}^2 x}{-\operatorname{sech}^2 x}$$ Cancel the negative signs, we get: $$\lim _{x \rightarrow \infty} \frac{\operatorname{csch}^2 x}{\operatorname{sech}^2 x}$$ We know that \(\operatorname{csch} x = \frac{1}{\sinh x}\) and \(\operatorname{sech} x = \frac{1}{\cosh x}\). Substitute these expressions into the limit: $$\lim _{x \rightarrow \infty} \frac{\left(\frac{1}{\sinh x}\right)^2} {\left(\frac{1}{\cosh x}\right)^2}$$ Simplify the equation: $$\lim _{x \rightarrow \infty} \frac{\cosh^2 x}{\sinh^2 x}$$

Evaluate the limit

We can rewrite the equation using the hyperbolic identity, \(\cosh^2 x - \sinh^2 x = 1\): $$\lim _{x \rightarrow \infty} \frac{\cosh^2 x}{\cosh^2 x - 1}$$ Now, divide both the numerator and denominator by \(\cosh^2 x\): $$\lim _{x \rightarrow \infty} \frac{1}{1 - \frac{1}{\cosh^2 x}}$$ As \(x\) approaches infinity, \(\frac{1}{\cosh^2 x}\) approaches zero, so: $$\lim _{x \rightarrow \infty} \frac{1}{1 - 0} = 1$$ Therefore, the limit of the given function is 1.

Key concepts

Limit Evaluation

Limit evaluation is a fundamental concept in calculus that allows us to understand the behavior of functions as they approach a specific point or infinity. In this case, we're looking at a limit as \(x\) approaches infinity. This can often result in indeterminate forms like \(\frac{0}{0}\), where both the numerator and the denominator approach zero. In such situations, l'Hôpital's Rule is a useful tool that allows us to simplify and evaluate the limit by differentiating the top and the bottom.
When you encounter such indeterminate forms, after confirming the form is \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), you apply l'Hôpital's Rule by taking derivatives of both the numerator and denominator until you eliminate the indeterminate form. If necessary, this process can be repeated. It's crucial to evaluate the limit of the new function after differentiation to ascertain the behavior of the original function as \(x\) approaches the particular value or infinity.

Hyperbolic Functions

Hyperbolic functions, like their trigonometric counterparts, include functions such as \(\sinh, \cosh, \tanh\), and \(\operatorname{coth}\). They are defined using exponential functions and have properties similar to trigonometric functions. For instance, \(\sinh(x) = \frac{e^x - e^{-x}}{2}\) and \(\cosh(x) = \frac{e^x + e^{-x}}{2}\), resembling how \(\sin(x)\) and \(\cos(x)\) are derived using the unit circle.
In the given problem, the hyperbolic functions involved are \(\operatorname{coth}(x)\) and \(\tanh(x)\). The hyperbolic cotangent, \(\operatorname{coth}(x) = \frac{\cosh(x)}{\sinh(x)}\), and hyperbolic tangent, \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\), play a crucial role in the differentiation process needed for applying l'Hôpital's Rule.
One key identity used to simplify the limit is \(\cosh^2(x) - \sinh^2(x) = 1\). This identity helps transform the expression into a more manageable form so the limit can be easily computed as \(x\) approaches infinity.

Derivatives

Derivatives are a powerful tool in calculus used to measure how a function changes as its input changes. For limit evaluation, particularly when using l'Hôpital's Rule, derivatives are necessary to convert an indeterminate form into a determinate one.
For hyperbolic functions, the derivatives are often simple, yet crucial. In our problem, the derivative of \(\operatorname{coth}(x)\) is \(-\operatorname{csch}^2(x)\), where \(\operatorname{csch}(x) = \frac{1}{\sinh(x)}\), and the derivative of \(\tanh(x)\) is \(\operatorname{sech}^2(x)\), with \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\).
These derivatives change how the original functions behave, particularly as \(x\) approaches infinity. In this case, they allow the application of l'Hôpital's Rule, simplifying the original expression into a form where the limit can be directly evaluated. Understanding how to differentiate these functions is vital in correctly applying the rule and reaching a meaningful conclusion to the limit problem.