Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} \frac{\cosh x-1}{x^{2}} $$

Short answer

The limit is \( \frac{1}{2} \).

Step-by-step solution

Identify the form

First, substitute \( x = 0 \) into the expression to determine the form. This gives us \( \frac{\cosh(0)-1}{0^2} = \frac{1-1}{0} = \frac{0}{0} \), which is an indeterminate form. Thus, l'Hôpital's Rule can be applied.

Apply l'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we differentiate the numerator and the denominator separately. The derivative of \( \cosh x \) is \( \sinh x \), and the derivative of \( x^2 \) is \( 2x \). Thus, the limit now becomes \( \lim _{x \to 0} \frac{\sinh x}{2x} \).

Re-evaluate the limit

Substitute \( x = 0 \) again into the new expression to check the form: \( \frac{\sinh(0)}{2(0)} = \frac{0}{0} \), which is still indeterminate. We apply l'Hôpital's Rule a second time.

Apply l'Hôpital’s Rule again

Differentiate the numerator and denominator again. The derivative of \( \sinh x \) is \( \cosh x \), and the derivative of \( 2x \) is \( 2 \). Now, the limit becomes \( \lim _{x \to 0} \frac{\cosh x}{2} \).

Calculate the limit

Finally, substitute \( x = 0 \) into the limit expression: \( \frac{\cosh(0)}{2} = \frac{1}{2} \), as \( \cosh(0) = 1 \). Therefore, the limit value is \( \frac{1}{2} \).

Key concepts

Indeterminate Forms

When calculating limits, you often encounter expressions that don’t initially make sense, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are known as **indeterminate forms**. They are called "indeterminate" because they do not directly reveal the limit of a function and need further analysis to determine the actual limit. Indeterminate forms arise frequently in calculus problems. For the given problem: \( \lim _{x \rightarrow 0} \frac{\cosh x-1}{x^{2}} \), when you substitute \( x = 0 \), you get \( \frac{0}{0} \). This is clearly indeterminate.

To resolve an indeterminate form, techniques such as factoring, approximation, or applying l'Hôpital's Rule are used. l'Hôpital's Rule states that if the limit of a ratio results in an indeterminate form, you can differentiate the numerator and denominator separately and then evaluate the limit again.

Hyperbolic Functions

Hyperbolic functions may seem mysterious, but they are just analogues of trigonometric functions. The most common hyperbolic functions include \( \sinh x \), \( \cosh x \), and \( \tanh x \). In simple terms, hyperbolic functions relate to the hyperbola much like how trigonometric functions relate to the circle.

For this exercise, you deal with the hyperbolic cosine function, \( \cosh x \), defined as \( \cosh x = \frac{e^x + e^{-x}}{2} \). It's useful to remember that \( \cosh 0 = 1 \), which helps in simplifying limits. The derivative of \( \cosh x \) is \( \sinh x \), making it convenient when applying l'Hôpital’s Rule.

Hyperbolic functions are key in many areas of mathematics that involve calculus and differential equations, as they often describe the shape of hanging cables and the catenary curve.

Derivatives

Derivatives are fundamental in determining the slope or rate of change of a function. When you encounter an indeterminate form in limits, finding the derivatives of the numerator and denominator can help resolve it, especially when applying l'Hôpital’s Rule

In the problem, you had to differentiate \( \cosh x \) and \( x^2 \) to simplify and evaluate the limit effectively. You find:

• The derivative of \( \cosh x \) is \( \sinh x \).
• The derivative of \( x^2 \) is \( 2x \).

After initially applying l'Hôpital’s Rule, you differentiate again because \( \frac{\sinh(0)}{2(0)} = \frac{0}{0} \) is still an indeterminate form. Then, the derivatives become \( \cosh x \) and \( 2 \), allowing you to substitute \( x = 0 \) and solve for \( \frac{1}{2} \).

Derivatives not only simplify limits but are also vital in optimization problems, motion analysis, and understanding the behavior of functions in calculus.