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Chapter 6: Problem 64

Which of the two limits exists? a. \(\lim _{x \rightarrow \infty} e^{3 x}\) b. \(\lim _{x \rightarrow \infty} e^{-3 x}\)

Short Answer

Expert verified
The limit \( \lim _{x \rightarrow \infty} e^{-3x} \) exists and is 0.

Step by step solution

01

Understand the Exponential Function Behavior

Before evaluating the limits, we need to recall the behavior of the exponential function. For a function of the form \( e^{ax} \): - If \( a > 0 \), as \( x \to \infty \), \( e^{ax} \to \infty \).- If \( a < 0 \), as \( x \to \infty \), \( e^{ax} \to 0 \).
02

Evaluate the First Limit

For the first limit, \( a = 3 > 0 \), which means:\[ \lim _{x \rightarrow \infty} e^{3x} = \infty \]Therefore, this limit does not exist as it approaches infinity.
03

Evaluate the Second Limit

For the second limit, \( a = -3 < 0 \), which means:\[ \lim _{x \rightarrow \infty} e^{-3x} = 0 \]This limit exists because it approaches a finite number, specifically 0.

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Key Concepts

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

Exponential Function Behavior
Exponential functions have a unique behavior that makes them both fascinating and crucial in mathematics and everyday applications. An exponential function can be identified by its form, \( e^{ax} \), where \( e \) is Euler's number (approximately 2.71828) and \( a \) is a constant.
  • If \( a > 0 \), the exponential function grows rapidly as \( x \) increases. This means as \( x \) becomes very large, the value of \( e^{ax} \) continues to increase without bound. This is known as exponential growth.
  • On the other hand, if \( a < 0 \), the exponential function decreases quickly as \( x \) increases. In this scenario, as \( x \) gets larger, \( e^{ax} \) gets closer and closer to zero. This is referred to as exponential decay.
Understanding these behaviors is crucial when evaluating limits involving exponential functions, particularly when examining how they act as \( x \) approaches infinity.
Infinite Limits
In mathematics, a limit describes how a function behaves as its input approaches a particular value. An infinite limit occurs when a function's value increases or decreases without bound as the input grows.For the exponential function \( e^{3x} \), where \( a = 3 > 0 \), as \( x \to \infty \), the function \( e^{3x} \) grows exponentially without limit. In simpler terms, the function shoots upward indefinitely as \( x \) becomes very large.
When we write \( \lim_{x \to \infty} e^{3x} = \infty \), we mean that there is no particular value it approaches鈥攈ence, the limit does not exist in a conventional sense because it doesn't settle on a finite number.
Finite Limits
Finite limits describe cases where as inputs of the function grow, the outputs stabilize at some specific value. Consider the exponential function \( e^{-3x} \), where \( a = -3 < 0 \). As \( x \to \infty \), this function behaves much differently than \( e^{3x} \). Instead of growing indefinitely, \( e^{-3x} \) declines and gets progressively closer to zero.
In this case, we can say that \( \lim_{x \to \infty} e^{-3x} = 0 \). Here, \( 0 \) is a definite magnitude, making it a finite limit. So, the function asymptotically approaches zero as \( x \) becomes larger, meaning the limit exists and is equal to zero.

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Most popular questions from this chapter

A telephone company estimates the maximum market for car phones in a city to be 10,000 . Total sales are proportional to both the number already sold and the size of the remaining market. If 100 phones have been sold at time \(t=0\) and after 6 months 2000 have been sold, find a formula for the total sales after \(t\) months. Use your answer to estimate the total sales at the end of the first year.

Manufacturers estimate the upper limit for sales of digital cameras to be 25 million annually and find that sales increase in proportion to both current sales and the difference between the sales and the upper limit. In 2005 sales were 6 million, and in 2008 were 20 million. Find a formula for the annual sales (in millions) \(t\) years after 2005 . Use your answer to predict sales in \(2012 .\)

Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. \(y^{\prime}=5 y(100-y)\) \(y(0)=10\)

Why do larger sized raindrops fall faster than smaller ones? It depends on the resistance they encounter as they fall through the air. For large raindrops, the resistance to gravity's acceleration is proportional to the square of the velocity, whereas for small droplets, the resistance is proportional to the first power of the velocity. More precisely, their velocities obey the following differential equations, with each differential equation leading to a different terminal velocity for the raindrop: i. \(\frac{d v}{d t}=32.2-0.1115 v^{2}\) ii. \(\frac{d v}{d t}=32.2-52.6 v\) iii. \(\frac{d v}{d t}=32.2-5260 v\) a. Use a slope field program to graph the slope field of differential equation (i) on the window \([0,3]\) by \([0,20]\) (using \(x\) and \(y\) instead of \(t\) and \(v\) ). From the slope field, must the solution curves rising from the bottom level off at a particular \(y\) -value? Estimate the value. This number is the terminal velocity (in feet per second) for a downpour droplet. b. Do the same for differential equation (ii), but on the window \([0,0.1]\) by \([0,1]\). What is the terminal velocity for a drizzle droplet? c. Do the same for differential equation (iii), but on the window \([0,0.001]\) by \([0,0.01]\). What is the terminal velocity for a fog droplet? d. At this speed [from part (c)], how long would it take a fog droplet to fall 1 foot? This shows why fog clears so slowly.

Find the area between the curve \(y=1 / x^{n}\) (for \(n>1\) ) and the \(x\) -axis from \(x=1\) to \(\infty\).
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