Problem 81 - Calculus for Biology and Medicine

Question

Use the limit laws to determine \(\lim _{n \rightarrow \infty} a_{n}=a .\) $$ \lim _{n \rightarrow \infty} \frac{n+2^{-n}}{n} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Recognize the Structure of the Sequence

The given sequence is \( \frac{n+2^{-n}}{n} \). We can split this expression into two fractions: \( \frac{n}{n} \) and \( \frac{2^{-n}}{n} \). This simplification will help us use limit laws effectively.

Step 2: Apply Limit Law to First Fraction

Evaluate the limit of the first fraction, \( \frac{n}{n} \), which simplifies to 1. Therefore, \( \lim_{n \to \infty} \frac{n}{n} = 1 \).

Step 3: Apply Limit Law to Second Fraction

For the second fraction, \( \lim_{n \to \infty} \frac{2^{-n}}{n} \), note that as \( n \to \infty \), \( 2^{-n} \) approaches 0 because exponential decay is faster than linear growth. Therefore, \( \lim_{n \to \infty} \frac{2^{-n}}{n} = 0 \).

Step 4: Combine Results Using Limit Laws

Using the sum of limits law, combine the results from Steps 2 and 3: \( \lim_{n \to \infty} \left( \frac{n}{n} + \frac{2^{-n}}{n} \right) = 1 + 0 = 1 \).

Key Concepts

Infinity Limits

When we talk about infinity limits, we're diving into the behavior of sequences or functions as they move towards infinity—forever and ever! More specifically, we're interested in the value that a sequence approaches as the index, usually denoted by \( n \), becomes infinitely large. In our exercise, we're asked to find the limit of the sequence \( \frac{n + 2^{-n}}{n} \) as \( n \to \infty \).
Why is this important? Understanding infinity limits helps us grasp the long-term trend or the 'destination' of mathematical expressions. Essentially, it helps us see what happens to an expression when the variable grows very, very large.
By evaluating \( \lim_{n \to \infty} \frac{n}{n} \), which equals 1, and \( \lim_{n \to \infty} \frac{2^{-n}}{n} \), which equals 0, we find that together they approach a limit of 1 over infinite values.

Exponential Decay in Sequences

Exponential decay refers to the process where a quantity decreases at a rate proportional to its current value. In mathematics, this often occurs in sequences and can significantly impact how a sequence's terms behave as the index increases.
In our exercise, part of the sequence includes the term \( 2^{-n} \). As \( n \) increases, \( 2^{-n} \) shrinks rapidly towards zero, because the exponential function \( 2^{-n} \) gets smaller much faster than \( n \) can grow. This is a key characteristic of exponential decay—and why it's important in determining the limit.

• The behavior is dictated by the negative exponent which causes the term to approach zero as the exponent increases.
• This means that even when it's divided by \( n \), which is growing linearly, \( \frac{2^{-n}}{n} \) approaches zero.

Overall, understanding exponential decay helps us resolve how seemingly complex behaviors, like those in sequences, resolve into simple outcomes such as zero.

Sum of Limits Law

The sum of limits law is a powerful tool in calculus that allows us to simplify the process of finding limits for complex expressions. It states that if you have two limits that exist, you can simply add them together if they are part of a combined equation.
In our problem, we have two parts, \( \lim_{n \to \infty} \frac{n}{n} \) and \( \lim_{n \to \infty} \frac{2^{-n}}{n} \). Calculating these independently gives us limits of 1 and 0, respectively. When adding these results based on the sum of limits law, we get:

\[ \lim_{n \to \infty} \left( \frac{n}{n} + \frac{2^{-n}}{n} \right) = 1 + 0 = 1 \]

• This simplification is essential for dealing with functions or sequences that have more than one term.
• It helps to break them down into smaller, more manageable parts that are easier to analyze and understand.

This law shows that we can focus on the behavior of separate components, then combine the results to find the overall limit of a sequence as it approaches infinity.