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L'Hôpital loophole Explain why I'Hôpital's Rule fails when applied to the limit \(\lim _{x \rightarrow \infty} \frac{\sinh x}{\cosh x}\) and then find the limit another way.

Short Answer

Expert verified
Answer: L'Hôpital's Rule fails in this case because applying it leads to an undefined limit due to continued oscillation, which doesn't help us find the actual limit. Instead, using an alternative method based on properties of hyperbolic trigonometric functions, we can determine that the limit of the function as \(x\) approaches infinity is 1.

Step by step solution

01

Check If L'Hôpital's Rule Is Applicable

Let us evaluate the limit of the given function as \(x\) approaches infinity:$$\lim_{x\to\infty} \frac{\sinh{x}}{\cosh{x}}$$To apply L'Hôpital's Rule, we need to ensure that the limit results in an indeterminate form such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Let us check the limits of the numerator and the denominator:$$\lim_{x\to\infty}\sinh{x}=\infty,\quad \lim_{x\to\infty}\cosh{x}=\infty$$The function's limit results in the indeterminate form \(\frac{\infty}{\infty}\). Therefore, it seems we can use L'Hôpital's Rule.
02

Apply L'Hôpital's Rule

Now, let's apply L'Hôpital's Rule and differentiate the numerator and the denominator with respect to \(x\). We get:$$\frac{d(\sinh{x})}{dx} = \cosh{x} , \quad \frac{d(\cosh{x})}{dx} = \sinh{x}$$Now applying L'Hospital's rule, we would get:$$\lim_{x\to\infty}\frac{\sinh x}{\cosh x}=\lim_{x\to\infty}\frac{cosh x}{sinh x}$$
03

Recognize The Limit

We notice that the limit after applying L'Hôpital's Rule is undefined due to continued oscillation. This means that applying L'Hôpital's Rule did not help us in finding the limit. We need to use an alternative method.
04

Alternative Method - Properties of Hyperbolic Trig Functions

We know that the hyperbolic sine and cosine functions can be defined as follows: $$\sinh{x}=\frac{e^x - e^{-x}}{2}, \quad \cosh{x}=\frac{e^x + e^{-x}}{2}$$We can rewrite the given function as:$$\frac{\sinh{x}}{\cosh{x}} = \frac{\frac{e^x - e^{-x}}{2}}{\frac{e^x + e^{-x}}{2}}$$Simplifying the function, we get:$$f(x)=\frac{e^x - e^{-x}}{e^x + e^{-x}}$$ Now, let's analyze the limit of this function as \(x \to \infty\):$$\lim_{x\to\infty}\frac{e^x - e^{-x}}{e^x + e^{-x}}$$ As \(x\) goes to infinity, the term \(e^{-x}\) becomes negligibly small in comparison to \(e^x\). Therefore, we can approximate the function as: $$f(x)\approx \frac{e^x}{e^x}$$
05

Find the limit

Now, we can find the limit of this approximated function:$$\lim_{x\to\infty}\frac{e^x}{e^x} = 1$$This is the reason why L'Hôpital's Rule fails when applied to the given limit. By using an alternative method based on properties of hyperbolic trigonometric functions, we found that the limit of the function as \(x\) approaches infinity is 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Limits
Limits are fundamental in calculus, acting as the bedrock for concepts such as derivatives and integrals. In essence, limits help us understand the behavior of a function as it approaches a specific point or value. When we say \( \lim_{x \to a} f(x) = L \), we mean that as \( x \) gets closer and closer to \( a \), the function \( f(x) \) approaches the value \( L \).

For example, considering the limit \( \lim_{x \to \infty} \frac{\sinh x}{\cosh x} \), we try to understand what value the function heads toward as \( x \) becomes very large. Here, techniques like L'Hôpital's Rule can sometimes aid us in resolving tricky limits, especially those that result in indeterminate forms. However, as demonstrated in our problem, this rule isn't always applicable, and thus understanding the functional behavior through an alternative method becomes essential.
  • Limits can approach a number or even infinity.
  • They help in evaluating the behavior of functions at points which are not necessarily defined.
  • Crucial to understanding continuous change and analysis in mathematics.
Grasping Indeterminate Forms
Indeterminate forms are a special category of limits that do not initially provide enough information to determine the limit's actual value. When evaluating a limit, encountering forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) is a sign of indeterminacy.

In our exercise, as \( x \to \infty \), both \( \sinh x \) and \( \cosh x \) go to infinity, leading to the \( \frac{\infty}{\infty} \) form. Even though L'Hôpital's Rule is meant to handle such cases by differentiating the numerator and the denominator, it didn’t resolve the limit here, leading us back to the original undetermined state.
  • Indeterminate forms show that more analysis is needed.
  • They often indicate complex functional behavior around a point.
  • L'Hôpital's Rule is a tool specifically designed to tackle these forms.
Exploring Hyperbolic Functions
Hyperbolic functions, denoted as \( \sinh \), \( \cosh \), and others, are analogs of the trigonometric functions but related to hyperbolas instead of circles. They are defined using exponential functions and have uses in various mathematical fields.

For the function \( \sinh x = \frac{e^x - e^{-x}}{2} \) and \( \cosh x = \frac{e^x + e^{-x}}{2} \), you can observe they behave vastly different from their trigonometric counterparts by growing exponentially as opposed to oscillating. This characteristic allows an altered approach to solving limits through the dominance of terms. In our specific problem, we leveraged the property that as \( x \to \infty \), \( e^{-x} \) becomes insignificant compared to \( e^x \), simplifying our limit to \( \frac{e^x}{e^x} \), which yields 1.
  • Hyperbolic functions arise in practical areas like engineering and physics.
  • They grow exponentially and are not periodic like trigonometric functions.
  • Useful in solving hyperbolic angle problems and describing hyperbolic geometry.

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