Question

For those sequences that converge, find the limit \(\lim _{n \rightarrow \infty} a_{n}\) a. \(a_{n}=\frac{n^{2}+1}{n^{3}+1} .\) b. \(a_{n}=\frac{3 n+1}{n+2}\). c. \(a_{n}=\left(\frac{3}{n}\right)^{1 / n}\). d. \(a_{n}=\frac{2 n^{2}+4 n^{3}}{n^{3}+5 \sqrt{2+n^{6}}} .\) e. \(a_{n}=n \ln \left(1+\frac{1}{n}\right)\). f. \(a_{n}=n \sin \left(\frac{1}{n}\right)\). g. \(a_{n}=\frac{(2 n+3) !}{(n+1) !} .\)

Short answer

The limits are a) 0, b) 3, c) \(e^{-1}\), d) 4, e) 1, f) 1. The sequence g does not have a finite limit.

Step-by-step solution

Sequence a

The first sequence is \(a_{n}=\frac{n^{2}+1}{n^{3}+1}\). Using L'Hopital's rule to solve this indeterminate form \(\frac{0}{0}\) we obtain \(\lim_{n \rightarrow \infty} \frac{2n}{3n^2} = 0\).

Sequence b

The second sequence is \(a_{n}=\frac{3n+1}{n+2}\). Again, using L'Hopital's rule for the form \(\infty/\infty\) we obtain \(\lim_{n \rightarrow \infty} \frac{3}{1} = 3\).

Sequence c

The third sequence is \(a_{n}=\left(\frac{3}{n}\right)^{1/n}\). Take the natural log of the sequence to simplify, then apply L'Hopital's rule to obtain that it converges to \(e^{-1}\)

Sequence d

The fourth sequence is \(a_{n}=\frac{2n^2+4n^3}{n^3+5\sqrt{2+n^6}}\). By dividing the numerator and the denominator by \(n^3\) we get the limit to be 4.

Sequence e

The fifth sequence is \(a_{n}=n\ln(1+\frac{1}{n})\). By replacing \(\frac{1}{n}\) with \(x\), we can simplify the sequence to the form that the limit of is computed to be 1.

Sequence f

The sixth sequence is \(a_{n}=n\sin(\frac{1}{n})\). If we replace \(\frac{1}{n}\) with \(x\), this simplifies to \(x\sin(x)\), as \(x\) approaches 0. The limit is 1.

Sequence g

The seventh sequence is \(a_{n} = \frac{(2n+3)!}{(n+1)!}\) which clearly goes to \(\infty\) as \(n\) increases, therefore it doesn't have a finite limit.

Key concepts

L'Hopital's Rule

When dealing with calculus problems, especially limits, you might encounter indeterminate forms, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms can be tricky, but L'Hopital's Rule provides a handy tool for simplifying them. This rule applies when you need to find the limit of a quotient of two functions, and both the numerator and denominator approach zero or infinity as \(x\) approaches a particular value.
L'Hopital's Rule states that if the limits of these functions lead to such indeterminate forms, you can differentiate the numerator and the denominator separately and then re-evaluate the limit. For example:

• If \(\lim_{x \to c} f(x) = 0\) and \(\lim_{x \to c} g(x) = 0\),
• or \(\lim_{x \to c} f(x) = \infty\) and \(\lim_{x \to c} g(x) = \infty\), then
• \(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\),

provided these derivatives keep giving a definitive limit after application.
This means you repeat the differentiation until you can calculate the limit, as seen in the sequence solutions for problems a, b, and c from the original exercise.

Convergence of Sequences

In mathematical analysis, the concept of the convergence of sequences is fundamental. A sequence \(a_n\) is said to converge to a limit \(L\) if, as \(n\) becomes indefinitely large, \(a_n\) approaches \(L\) as closely as we like. Mathematically, for every given positive number \(\varepsilon\), there exists a natural number \(N\) such that for all \(n > N\), \(|a_n - L| < \varepsilon\).
This means no matter how small the \(\varepsilon\), a point in the sequence will eventually become so close to \(L\) that it stays within this \(\varepsilon\) distance. For example, in sequences like \(b\), \(b_n = \frac{3n+1}{n+2}\), as \(n\) grows, the sequence approaches 3, making 3 the limit.
Understanding convergence is crucial in calculus and analysis, as it often arises in the context of series, and differentiable functions.

Indeterminate Forms

When evaluating the limits of functions, sometimes the direct substitution of values into a function yields forms that are "indeterminate" like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), and many others. These forms don't provide sufficient information to determine the actual limit.
This is where you need techniques like L'Hopital's Rule or algebraic manipulation to resolve these ambiguities. Indeterminate forms arise mainly because the endpoint is not well-defined without additional analysis. For example, an indeterminate form like \(\frac{0}{0}\) can occur with expressions like in sequence a from the original exercise, where both the numerator and the denominator tend to zero.
In such cases, solving these forms involves simplifying the function or using derivatives, which help reveal the true behavior as the variable approaches the critical point.