Question

Use l'Hôpital's Rule to evaluate the following limits. $$\underset{x\to \mathrm{\infty }}{lim}\frac{1-\coth x}{1-\tanh x}$$

Short answer

Question: Determine the limit of the given function as x approaches infinity: $$\underset{x\to \mathrm{\infty }}{lim}\frac{1-\coth x}{1-\tanh x}$$ Answer: The limit of the given function as x approaches infinity is 1.

Step-by-step solution

Determine if the limit is an indeterminate form

As x approaches infinity, we have: 1. $\coth (x)=\frac{\cosh (x)}{\sinh (x)}$ approaches 1. This is because for large x, $\cosh (x)$ and $\sinh (x)$ have the same exponential growth, so their ratio approaches 1. 2. $\tanh (x)=\frac{\sinh (x)}{\cosh (x)}$ approaches 1. This is because for large x, $\sinh (x)$ and $\cosh (x)$ have the same exponential growth, so their ratios approach 1. Therefore, as x approaches infinity, the given function approaches the indeterminate form $\frac{1-1}{1-1}=\frac{0}{0}$.

Apply l'Hôpital's Rule

Since the limit is an indeterminate form, we can apply l'Hôpital's Rule, which states: $$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$$ If the limit on the right-hand side exists or is infinite. To do this, we need to first find the derivatives of the numerator and denominator functions: 1. For the numerator: $f(x)=1-\coth (x)$. By chain rule, we have $f^{\prime }(x)=0-\frac{d(\coth (x))}{dx}=-\frac{1}{{\sinh }^2(x)}$. 2. For the denominator: $g(x)=1-\tanh (x)$. By chain rule, we have $g^{\prime }(x)=0-\frac{d(\tanh (x))}{dx}=-{\mathrm{sech}}^2(x)$. Now, we substitute these derivatives back into the limit function and evaluate the limit: $$\underset{x\to \mathrm{\infty }}{lim}\frac{-\frac{1}{{\sinh }^2(x)}}{-{\mathrm{sech}}^2(x)}$$

Simplify the expression and evaluate the limit

Now, we simplify this expression by changing sech^2(x) into its equivalent expression in terms of cosh: $$-{\mathrm{sech}}^2(x)=-\frac{1}{{\cosh }^2(x)}$$ The limit expression now becomes: $$\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{{\sinh }^2(x)}}{\frac{1}{{\cosh }^2(x)}}$$ By simplification, the expression is now: $$\underset{x\to \mathrm{\infty }}{lim}\frac{{\cosh }^2(x)}{{\sinh }^2(x)}$$ Now using the identity ${\cosh }^2(x)-{\sinh }^2(x)=1$, we rewrite the expression as: $$\underset{x\to \mathrm{\infty }}{lim}\frac{{\sinh }^2(x)+1}{{\sinh }^2(x)}$$ Now, we can rewrite the expression in terms of the hyperbolic functions themselves: $$\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{{\sinh }^2(x)})$$ Since $\frac{1}{{\sinh }^2(x)}$ goes to 0 as x approaches infinity, the limit converges to 1 + 0.

Write the final result

Thus, the limit of the given function as x approaches infinity is: $$\underset{x\to \mathrm{\infty }}{lim}\frac{1-\coth x}{1-\tanh x}=1$$

Key concepts

Indeterminate Forms

When dealing with limits in calculus, particularly as variables approach infinity or other specific values, we may encounter expressions that result in an indeterminate form. The expression $\frac{0}{0}$ is a classic example of such an indeterminate form. Indeterminate forms arise when the contributions of the numerator and denominator tend to cancel each other out, making direct evaluation impossible. This happens because both the numerator and the denominator individually approach zero, rendering the limit uncertain and undefined without further analysis.

In practical terms, knowing that a limit expression like $\frac{f(x)}{g(x)}$ turns into $\frac{0}{0}$ as $x\to a$, suggests the utilization of techniques like factoring, algebraic manipulation, or more commonly, l'Hôpital's Rule. These methods help in resolving the ambiguity of the indeterminate form, allowing us to find a precise limit as in the given exercise. By differentiating both the numerator and the denominator, l'Hôpital's Rule offers a way to overcome the obstacle posed by indeterminacy.

Hyperbolic Functions

Hyperbolic functions have similar properties to trigonometric functions but are defined via exponential functions. Let's look at the characteristics of the hyperbolic functions involved in this limit problem:

• Coth function: The hyperbolic cotangent function, $\coth (x)=\frac{\cosh (x)}{\sinh (x)}$, is defined as the ratio of the hyperbolic cosine and hyperbolic sine functions. As $x$ approaches infinity, both the hyperbolic sine and cosine functions grow exponentially, thus, their ratio tends towards 1.


• Tanh function: Similarly, the hyperbolic tangent function, $\tanh (x)=\frac{\sinh (x)}{\cosh (x)}$, also simplifies to 1 as $x$ goes to infinity due to the equivalent exponential growth of the hyperbolic sine and cosine functions.

These functions play a crucial role in limit problems and exhibit behaviors key for evaluating the given limit expression effectively. Understanding how these functions behave at extreme values helps predict their limits in such calculus problems.

Calculus Limits

In calculus, finding the limit of a function as an input approaches a certain value is a fundamental concept. Limits help us understand the behavior of functions in situations where values approach a point but do not necessarily equal that point.

This is particularly evident when exploring the limits of rational functions made up of complex expressions, such as the combination of hyperbolic functions we deal with in the given exercise.

L'Hôpital's Rule, a critical tool for handling indeterminate forms, allows us to compute limits by substituting derivatives into the original limit expression. With calculus limits, we're often verifying how a function behaves asymptotically - that is, describing the end behavior of functions as inputs grow large or small without bounds. By transforming a limit problem involving hyperbolic functions into a simpler mathematical form, we can more easily determine the behavior and actual limit as one observes under extreme input conditions. This is precisely what occurred in the solution to the given exercise, leading to the conclusion that at infinity, the function resolves to an elegant limit of 1.