Question

Use Theorem 4.1.7 to find \(\lim _{n \rightarrow \infty} s_{n}\). (a) \(s_{n}=\frac{\alpha+n}{\beta+n} \quad(\beta>0)\) (b) \(s_{n}=\cos \frac{1}{n}\) (c) \(s_{n}=n \sin \frac{1}{n}\) (d) \(s_{n}=\log n-n\) (e) \(s_{n}=\log (n+1)-\log (n-1)\)

Short answer

Question: Determine the limit of each of the following sequences: 1. \(\lim_{n \rightarrow \infty} \frac{\alpha+n}{\beta+n}\) Answer: The limit of the sequence is 1. 2. \(\lim_{n \rightarrow \infty} \cos{\frac{1}{n}}\) Answer: The limit of the sequence is 1. 3. \(\lim_{n \rightarrow \infty} n\sin\frac{1}{n}\) Answer: The limit of the sequence is 1. 4. \(\lim_{n \rightarrow \infty} \log{n} - n\) Answer: The limit of the sequence is -∞. 5. \(\lim_{n \rightarrow \infty} \log\left(n+1\right)-\log\left(n-1\right)\) Answer: The limit of the sequence is 0.

Step-by-step solution

1. \(\lim_{n \rightarrow \infty} \frac{\alpha+n}{\beta+n}\)

Continuous expression: \(s(x) = \frac{\alpha+x}{\beta+x}\) Now find the limit as x approaches infinity: \(\lim_{x \rightarrow \infty} \frac{\alpha+x}{\beta+x} = \lim_{x \rightarrow \infty} \frac{1+\frac{\alpha}{x}}{1+\frac{\beta}{x}} = \frac{1+0}{1+0} = 1\)

2. \(\lim_{n \rightarrow \infty} \cos{\frac{1}{n}}\)

Continuous expression: \(s(x) = \cos{\frac{1}{x}}\) Now find the limit as x approaches infinity: \(\lim_{x \rightarrow \infty} \cos{\frac{1}{x}} = \cos{0}=1\)

3. \(\lim_{n \rightarrow \infty} n\sin\frac{1}{n}\)

Continuous expression: \(s(x) = x\sin\frac{1}{x}\) Now find the limit as x approaches infinity: Using L'Hopital's rule since \(\frac{\sin\frac{1}{x}}{1/x}\) is an indeterminate form: \(\lim_{x \rightarrow \infty} x\sin\frac{1}{x} = \lim_{x \rightarrow \infty} \frac{\sin\frac{1}{x}}{\frac{1}{x}} = \lim_{x \rightarrow \infty} \frac{\cos\frac{1}{x}\cdot\frac{-1}{x^2}}{-\frac{1}{x^2}} = \lim_{x \rightarrow \infty} \cos\frac{1}{x} = \cos{0} = 1\)

4. \(\lim_{n \rightarrow \infty} \log{n} - n\)

Continuous expression: \(s(x) = \log{x} - x\) Now find the limit as x approaches infinity: \(\lim_{x \rightarrow \infty}\left( \log{x} - x\right)\). Since exponential growth outpaces logarithmic growth, the function goes to negative infinity as \(x\) goes to infinity. Therefore: \(\lim_{x \rightarrow \infty} \log{x} - x = -\infty\)

5. \(\lim_{n \rightarrow \infty} \log\left(n+1\right)-\log\left(n-1\right)\)

Continuous expression: \(s(x) = \log\left(x+1\right) - \log\left(x-1\right)\). Using the properties of logarithms, we can rewrite the function as: \(s(x) = \log\frac{x+1}{x-1}\) Now find the limit as x approaches infinity: \(\lim_{x \rightarrow \infty} \log\frac{x+1}{x-1} = \lim_{x \rightarrow \infty} \log\frac{1+\frac{1}{x}}{1-\frac{1}{x}} = \log\frac{1}{1} = \log{1} = 0\)

Key concepts

Continuous Functions

In calculus, a function is considered continuous if there are no breaks, jumps, or holes at any point in its domain. A continuous function allows you to trace a graph without lifting your pen off the paper. This property is crucial when discussing limits because the limit behavior of a function at infinity or any other point is easier to determine if the function is continuous.
For example, in part (b) of the original exercise with the limit \(\lim_{n \rightarrow \infty} \cos{\frac{1}{n}}\), the function \(\cos(x)\) is continuous across its entire domain. Therefore, as \(n\) approaches infinity and \(\frac{1}{n}\) approaches zero, the entire cosine function will approach \(\cos(0)\), which is 1. This concept is applicable to any continuous trigonometric and logarithmic functions, enabling straightforward computation of limits.

L'Hôpital's Rule

L'Hôpital's Rule provides a technique to evaluate limits that result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule states that if the limits of both the numerator and the denominator of a fraction are 0 or infinity, the original limit can be evaluated using the limits of their derivatives.
For the exercise \(\lim_{n \rightarrow \infty} n\sin\frac{1}{n}\), we recognize an indeterminate form \(\frac{0}{0}\) when rewritten as \(\lim_{x \rightarrow \infty} \frac{\sin\frac{1}{x}}{\frac{1}{x}}\). Using L'Hôpital's Rule, differentiate the numerator and denominator, resulting in \(\cos\frac{1}{x}\) in the limit computation. Evaluating at infinity, \(\cos 0 = 1\). L'Hôpital's rule simplifies the complex indeterminate problems into manageable ones using derivatives.

Logarithmic Functions

Logarithmic functions, such as \(\log(x)\), are powerful tools in calculus due to their properties. They are the inverse of exponential functions and grow at a slower rate. This slow growth has significant implications when evaluating limits, particularly as \(x\) approaches infinity.
In part (d) of the exercise \(\lim_{n \rightarrow \infty} \log{n} - n\), the challenge lies in the behavior at infinity. While \(\log{n}\) increases, \(n\) increases much faster, overriding any growth by \(\log(x)\), leading the function to \(-\infty\). Furthermore, in part (e), the exercise \(\log(x+1) - \log(x-1)\) is simplified using properties of logarithms to evaluate limits more easily, arriving at 0 as \(x\) grows large, showing how logarithmic properties aid in limit evaluation.

Trigonometric Functions

Trigonometric functions like sine and cosine have periodic properties that are particularly useful in limit problems. The key trigonometric identities and their behaviors at specific limit points can offer a roadmap to easy solutions.
Specifically, in part (c) with \(\lim_{n \rightarrow \infty} n\sin\frac{1}{n}\), the trigonometric function \(\sin(x)\) is analyzed within a continuous and smooth curve. Recognizing the indeterminate form allows us to apply L'Hôpital’s rule, simplifying the expression and resulting in \(\cos(0) = 1\). Similarly, in part (b) with \(\lim_{n \rightarrow \infty} \cos\frac{1}{n}\), understanding that \(\cos(x)\) gently transitions between values helps infer its behavior as \(x\) approaches zero, simplifying to 1 due to continuity of cosine around zero.