se the formula, lim a" o, if a > 1 lim a" = 0, if 0 < a < 1

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Hi, thanks for answering this question. I have another question about this. For the part circled in green, where did this come from? Is there a name for this rule? I never saw this anywhere in my notes so it's throwing me off. 

The image contains a mathematical expression for educational purposes. The expression is written as follows:

36. 

\[ \left\{ \frac{e^{n/10}}{2^n} \right\} \]

There are small, hand-drawn shapes above and beside the number "36," resembling a loop or arc and a small circle, but they do not appear to relate directly to the mathematical expression.
Transcribed Image Text:The image contains a mathematical expression for educational purposes. The expression is written as follows: 36. \[ \left\{ \frac{e^{n/10}}{2^n} \right\} \] There are small, hand-drawn shapes above and beside the number "36," resembling a loop or arc and a small circle, but they do not appear to relate directly to the mathematical expression.
## Calculus Example: Exponential Limit

### Step 1

The given data is:

\[
e^{1/10} = \sqrt[10]{e} = 1.1
\]

\[
r = \frac{e^{1/10}}{2}, \quad 0 < r < 1
\]

To find the value of:

\[
\lim_{{n \to \infty}} \frac{e^{n/10}}{2^n}
\]

### Step 2

Use the formula:

\[
\lim_{{n \to \infty}} a^n = \infty, \quad \text{if } a > 1
\]

\[
\lim_{{n \to \infty}} a^n = 0, \quad \text{if } 0 < a < 1
\]

Now,

\[
\lim_{{n \to \infty}} \frac{e^{n/10}}{2^n} = \lim_{{n \to \infty}} r^n
\]

\[ 
= 0 \quad (\because 0 < r < 1) 
\]

This example demonstrates the application of limits to exponential functions, useful in evaluating the behavior of functions as \( n \) approaches infinity.
Transcribed Image Text:## Calculus Example: Exponential Limit ### Step 1 The given data is: \[ e^{1/10} = \sqrt[10]{e} = 1.1 \] \[ r = \frac{e^{1/10}}{2}, \quad 0 < r < 1 \] To find the value of: \[ \lim_{{n \to \infty}} \frac{e^{n/10}}{2^n} \] ### Step 2 Use the formula: \[ \lim_{{n \to \infty}} a^n = \infty, \quad \text{if } a > 1 \] \[ \lim_{{n \to \infty}} a^n = 0, \quad \text{if } 0 < a < 1 \] Now, \[ \lim_{{n \to \infty}} \frac{e^{n/10}}{2^n} = \lim_{{n \to \infty}} r^n \] \[ = 0 \quad (\because 0 < r < 1) \] This example demonstrates the application of limits to exponential functions, useful in evaluating the behavior of functions as \( n \) approaches infinity.
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