Problem 19 - Calculus of a Single Variable

Question

In Exercises 17–22, find the limit. $$ \lim _{x \rightarrow \infty} \operatorname{sech} x $$

Step-by-Step Solution

Answer

0

Step 1: Write sech function in terms of exponentials

First replace \(sech(x)\) with its equivalent exponential notation, which gives \(\frac{2}{e^x + e^{-x}}\).

Step 2: Simplify the function by removing e^-x

As \(x\) approaches infinity, \(e^{-x}\) gets very close to zero. So, we can ignore this part to simplify the equation, which gives \(\frac{2}{e^x + 0}\). Under these conditions, the expression is simplifying to \(\frac{2}{e^x}\).

Step 3: Find the limit

The limit of the function \(\frac{2}{e^x}\) as \(x\) approaches infinity is zero, because the numerator is a finite number and the denominator is an infinitely large number. Hence, the limit of \(sech(x)\) as \(x\) approaches infinity is 0.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the well-known trigonometric functions, but they are expressed using exponential functions. One such function is the hyperbolic secant, denoted as \( \operatorname{sech}(x) \). It is defined through exponential functions as \( \operatorname{sech}(x) = \frac{2}{e^x + e^{-x}} \).

These functions arise naturally in various contexts, particularly in areas like calculus and hyperbolic geometry. They have a close relationship with the regular trigonometric functions but behave differently, especially as you consider limits at infinity.

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
• \( \operatorname{sech}(x) = \frac{1}{\cosh(x)} = \frac{2}{e^x + e^{-x}} \)

Understanding hyperbolic functions is crucial as they provide solutions to some differential equations just like trigonometric functions do, and they even have applications in areas such as relativity and signal processing.

Exponential Functions

Exponential functions are a fundamental part of calculus, characterized by their growth rates. The general form of an exponential function is \( f(x) = a \cdot e^{bx} \), where \( e \) is the base of natural logarithms, approximately equal to 2.718.

An important property of exponential functions, like \( e^x \), is their rapid increase. As \( x \) approaches infinity, \( e^x \) grows without bound. This property is fundamental in calculus limits since the parts involving \( e^x \) can often dominate an expression when \( x \) becomes very large or very small. Conversely, \( e^{-x} \) decreases toward zero as \( x \) goes to infinity.

• If \( x \to \infty \), then \( e^x \to \infty \).
• If \( x \to \infty \), then \( e^{-x} \to 0 \).

This behaviour of exponential functions makes them extremely useful in understanding and evaluating limits and asymptotic behaviour in calculus.

Infinity in Calculus

In calculus, the concept of infinity is pivotal, especially when dealing with limits. It provides a way to comprehend values that grow unbounded or shrink to infinitesimally small.

When we say \( x \to \infty \), we're examining what happens to a function as \( x \) gets larger and larger. Understanding this concept is crucial in evaluating limits and helps in understanding the behaviour of functions at extreme values.

• Functions with finite numerators and infinite denominators tend toward zero.
• Functions with infinite growth in numerators can grow without bound, depending on the denominator.
• If both numerator and denominator grow infinitely, the limit might need special techniques like L'Hôpital's rule.

Limits involving infinity help us define asymptotic behaviour and are foundational for discerning trends and making predictions in complex calculus problems.