In the following problems, find the limit of the given sequence as \(n \rightarrow \infty\). $$\frac{(n+1)^{2}}{\sqrt{3+5 n^{2}+4 n^{4}}}$$

Short Answer

Expert verified
\( \frac{1}{2} \)

Step by step solution

01

Simplify the sequence

Identify the dominant terms in the numerator and the denominator to simplify the expression. In the numerator, \((n+1)^2\), the term \^2\ dominates as \(n \rightarrow \infty\). In the denominator, \( \sqrt{3 + 5n^2 + 4n^4}\), the term \4n^4\ dominates.
02

Extract dominant terms

Rewrite the sequence by considering the dominant terms. For the numerator, keep \^2\ and for the denominator, simplify \sqrt{4n^4}\. This gives: \[\frac{(n+1)^2}{\sqrt{4n^4}} \approx \frac{n^2+2n+1}{2n^2}\]
03

Simplify the expression

Simplify the fraction \frac{n^2 + 2n + 1}{2n^2}\ by dividing each term in the numerator by \^2\: \[\frac{n^2}{2n^2} + \frac{2n}{2n^2} + \frac{1}{2n^2} = \frac{1}{2} + \frac{1}{n} + \frac{1}{2n^2}\]
04

Find the Limit

As \(n \rightarrow \infty\), the terms \frac{1}{n}\ and \frac{1}{2n^2}\ approach 0. Therefore, the limit can be evaluated as: \[ \frac{1}{2} + 0 + 0 = \frac{1}{2} \]

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Key Concepts

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

Dominant Terms
When dealing with sequences, especially as variables approach infinity, identifying dominant terms is crucial.
Dominant terms are the parts of the expression that have the highest power or growth rate.
Let's consider the given sequence: \(\frac{(n+1)^{2}}{\text{sqrt}(3+5n^2+4n^4)}\).
In the numerator, the dominant part is \(n^2 \), because as \(n \rightarrow \text{infinity}\), \(n^2\) grows faster than \(2n\) and \(1\).
Similarly, in the denominator, the term \(4n^4\) dominates since it has the highest power.

Understanding these dominant terms allows us to simplify the expression more effectively by focusing on the major contributing terms.
This makes it easier to evaluate limits as \(n \rightarrow \text{infinity}\).
Simplifying Expressions
Simplifying expressions helps reduce complex forms into manageable terms.
Once we identify dominant terms, we extract and rewrite them.

For the numerator \( (n+1)^2\), which expands to \(n^2 + 2n + 1\), we focus on \(n^2\).
For the denominator \( \text{sqrt}(3 + 5n^2 + 4n^4)\), we focus on \( \text{sqrt}(4n^4) \), which simplifies to \(2n^2\).

Thus, our expression now is \(\frac{(n+1)^2}{\text{sqrt}(4n^4)} \), which simplifies to \( \frac{n^2 + 2n + 1}{2n^2}\).
We then simplify this fraction by dividing each term in the numerator by \(2n^2\):

\(\frac{n^2}{2n^2} + \frac{2n}{2n^2} + \frac{1}{2n^2}\).
This simplifies to \(\frac{1}{2} + \frac{1}{n} + \frac{1}{2n^2}\).

By simplifying, we make it easier to evaluate limits.
Evaluating Limits at Infinity
To evaluate the limit of a simplified sequence as \(n \rightarrow \text{infinity}\), observe the behavior of each term.

Consider our simplified expression: \(\frac{1}{2} + \frac{1}{n} + \frac{1}{2n^2}\).
As \(n\) approaches infinity, terms like \( \frac{1}{n}\) and \( \frac{1}{2n^2}\) approach 0.
This is because any number divided by an increasingly larger number tends to zero.

Thus, the expression simplifies further to:
\(\frac{1}{2} + 0 + 0 = \frac{1}{2}\).

By focusing on the dominant terms and simplifying expressions, you can accurately evaluate the limits of sequences at infinity.
Practicing these steps ensures a deeper understanding and ability to handle more complex problems.

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Most popular questions from this chapter

(a) It is clear that you (or your computer) can't find the sum of an infinite series just by adding up the terms one by one. For example, to get \(\zeta(1.1)=\) \(\sum_{n=1}^{\infty} 1 / n^{1.1}(\text { see Problem } 15.22)\) with error \(<0.005\) takes about \(10^{33}\) terms. To see a simple alternative (for a series of positive decreasing terms) look at Figures 6.1 and 6.2. Show that when you have summed \(N\) terms, the sum \(R_{N}\) of the rest of the series is between \(I_{N}=\int_{N}^{\infty} a_{n} d n\) and \(I_{N+1}=\int_{N+1}^{\infty} a_{n} d n\) (b) Find the integrals in (a) for the \(\zeta(1.1)\) series and verify the claimed number of terms needed for error \(<0.005 .\) Hint: Find \(N\) such that \(I_{N}=0.005 .\) Also find upper and lower bounds for \(\zeta(1.1)\) by computing \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N}^{\infty} n^{-1.1} d n\) and \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N+1}^{\infty} n^{-1.1} d n\) where \(N\) is far less than \(10^{33} .\) Hint: You want the difference between the upper and lower limits to be about \(0.005 ;\) find \(\mathrm{N}\) so that term \(a_{N}=0.005\).

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