Question

Evaluate the limit of the following sequences. $$a_{n}=\frac{6^{n}+3^{n}}{6^{n}+n^{100}}$$

Short answer

Answer: The limit of the sequence \(a_{n}\) is 1 as \(n\) approaches infinity.

Step-by-step solution

1. Identify the dominating terms

When determining the limit of a sequence as \(n\) approaches infinity, it's important to focus on the dominating terms in both the numerator and the denominator. As we can see, in the numerator, \(6^n\) will be the dominating term because it grows much faster than \(3^n\). Similarly, in the denominator, \(6^n\) dominates over \(n^{100}\), especially as \(n\) increases.

2. Factor out the dominating term

Since we identified that \(6^n\) is the dominating term in both the numerator and the denominator, we should factor it out from the expression. Doing so will make it easier to find the limit. Here is the expression after factoring out \(6^n\): $$a_{n}=\frac{6^{n}(1+\frac{3^n}{6^n})}{6^{n}(1+\frac{n^{100}}{6^n})}$$

3. Simplify the expression

Now that we have factored out \(6^{n}\), we can simplify the expression further. We see that the \(6^n\) terms in the numerator and denominator cancel each other out. Furthermore, we can rewrite the exponent using properties of exponents: $$a_{n}=\frac{1+\frac{3^n}{6^n}}{1+\frac{n^{100}}{6^n}} = \frac{1+\frac{3^n}{(2\cdot3)^n}}{1+\frac{n^{100}}{(2\cdot3)^n}} = \frac{1+\frac{1}{2^n}}{1+\frac{n^{100}}{2^n \cdot 3^n}}$$

4. Find the limit

With the simplified expression, we can now find the limit of the sequence as \(n\) approaches infinity. As \(n\) goes to infinity, both \(\frac{1}{2^n}\) and \(\frac{n^{100}}{2^n \cdot 3^n}\) approach 0, because the denominator grows much faster than the numerator. Therefore, we have: $$\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}\frac{1+\frac{1}{2^n}}{1+\frac{n^{100}}{2^n \cdot 3^n}} = \frac{1+0}{1+0} = \boxed{1}$$ So, the limit of the given sequence \(a_{n}\) is 1 as \(n\) approaches infinity.

Key concepts

Understanding Dominant Terms in Sequences

When dealing with sequences, identifying the dominant term is crucial for determining the behavior as the sequence approaches infinity. A dominant term is the term in the sequence that grows significantly faster than others when the variable, often denoted as \( n \), becomes very large.
For example, in the sequence \( a_{n}=\frac{6^{n}+3^{n}}{6^{n}+n^{100}} \), we have expressions in both the numerator and the denominator. To simplify and find limits, compare how different terms grow:

• In the numerator: \( 6^n \) grows faster than \( 3^n \).
• In the denominator: \( 6^n \) also dominates over \( n^{100} \).

Why is this important? Because focusing on the dominant terms helps simplify the sequence and makes it easier to understand.
This simplification step allows you to factor out these dominant terms and isolate the significant components affecting the growth rate of the sequence.

Factoring in Calculus: Simplifying Expressions

Factoring is a powerful technique in calculus that allows us to simplify expressions, especially when evaluating limits of sequences or functions. By factoring out the dominant term, we can often remove complexity and directly observe how the sequence behaves. Let’s see how this works in the given sequence:
After identifying \( 6^n \) as the dominant term:

• Factor it out from both the numerator and the denominator: \( a_{n}=\frac{6^{n}(1+\frac{3^n}{6^n})}{6^{n}(1+\frac{n^{100}}{6^n})} \).
• Cancelling \( 6^n \) simplifies the sequence to \( \frac{1+\frac{3^n}{6^n}}{1+\frac{n^{100}}{6^n}} \).

This expression is much more manageable. Factoring helps us reduce the problem to a simpler form where we can more easily analyze the limit behavior.
Remember, factoring in calculus is not just a mechanical step. It is about revealing the significant components of the expression, which helps in understanding the mathematical behavior of sequences as they grow large.

Exploring Exponents and Limits

Understanding the relationship between exponents and limits is key to evaluating the behavior of sequences. Exponents indicate how rapidly terms in a sequence grow, and this understanding can be leveraged to evaluate limits efficiently. Let’s apply this to the exercise at hand:
In the simplified sequence expression, \( \frac{1+\frac{3^n}{(2\cdot3)^n}}{1+\frac{n^{100}}{(2\cdot3)^n}} \):

• The terms \( \frac{1}{2^n} \) and \( \frac{n^{100}}{2^n \cdot 3^n} \) each go to 0 as \( n \rightarrow \infty \) because these terms involve exponential growth in the denominator.

Exponential growth occurs when the base of an exponent is raised to a power. In this example, both terms in the expression decrease toward 0 because their exponential growth (2^n and (2\cdot3)^n) in the denominator outpaces the numerator.
This behavior is fundamental when calculating limits, as it helps simplify expressions to their essence — which often result in clear and simple limits, like in this case, 1. Thus, understanding exponents is crucial in deciding how individual terms in a sequence contribute to its limit.