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Chapter 1: Problem 29

$\lim _{\mathrm{n} \rightarrow \infty}\left(\frac{\mathrm{n}}{\mathrm{n}^{2}-2}+\frac{4^{\mathrm{n}}(-1)^{\mathrm{n}}}{2^{\mathrm{n}}-1}\right)^{-1}$ is equal to (A) 2 (B) (C) 0 (D) None of these

Short Answer

Expert verified
Answer: The value of the limit is 1, which corresponds to option (D), none of the given options.

Step by step solution

01

Break the expression into two limits

We can separate the limit of the sum of two functions into the sum of two separate limits: $$ \lim _{n \rightarrow \infty}\left(\frac{n}{n^{2}-2}+\frac{4^{n}(-1)^{n}}{2^{n}-1}\right)^{-1} = \left( \lim _{n \rightarrow \infty} \frac{n}{n^{2}-2} + \lim _{n \rightarrow \infty} \frac{4^{n}(-1)^{n}}{2^{n}-1} \right) ^{-1}. $$
02

Simplify the first limit

To simplify the first limit \(\lim _{n \rightarrow \infty}\frac{n}{n^{2}-2}\), we can divide both the numerator and the denominator by \(n^2\): $$ \lim _{n \rightarrow \infty}\frac{n}{n^{2}-2} = \lim _{n \rightarrow \infty}\frac{\frac{n}{n^2}}{\frac{n^2-2}{n^2}} = \lim _{n \rightarrow \infty}\frac{1}{1-\frac{2}{n^2}}. $$ As \(n\) approaches infinity, \(-2/n^2\) approaches 0. Thus, the first limit converges to: $$ \lim _{n \rightarrow \infty}\frac{1}{1-\frac{2}{n^2}} = \frac{1}{1} = 1. $$
03

Simplify the second limit

First, we can use the property \(4^n = (2^2)^n = 2^{2n}\) to rewrite the second limit as: $$ \lim _{n \rightarrow \infty}\frac{4^{n}(-1)^{n}}{2^{n}-1} = \lim _{n \rightarrow \infty}\frac{2^{2n}(-1)^{n}}{2^{n}-1}. $$ Now, let's divide both the numerator and the denominator by \(2^n\), getting: $$ \lim _{n \rightarrow \infty}\frac{2^{2n}(-1)^{n}}{2^{n}-1} = \lim _{n \rightarrow \infty}\frac{2^{n}(-1)^{n}}{1-\frac{1}{2^{n}}}. $$ As \(n\) approaches infinity, \(-1/n\) oscillates between -1 and 1, creating an alternating sequence and making the second limit difficult to evaluate directly. Therefore, we can apply the Squeeze Theorem.
04

Use the Squeeze Theorem to handle the second limit

Let's consider the sequence \(a_n = 2^n(-1)^n\). We know that \(-2^n \leq a_n \leq 2^n\) for any \(n\). Thus, we can write: $$ -1 \leq \frac{a_n}{2^n} \leq 1. $$ As \(n\) approaches infinity, both the lower and upper bounds approach 0: $$ \lim _{n \rightarrow \infty} -1 = \lim _{n \rightarrow \infty} 1 = 0. $$ By the Squeeze Theorem, we have: $$ \lim _{n \rightarrow \infty}\frac{2^{n}(-1)^{n}}{2^{n}} = 0. $$ Therefore, the second limit also converges to 0.
05

Combine the two limits

Now, let's substitute the values of the two limits back into the original expression: $$ \left(\lim _{n \rightarrow \infty} \frac{n}{n^{2}-2} + \lim _{n \rightarrow \infty} \frac{4^{n}(-1)^{n}}{2^{n}-1}\right)^{-1} = (1 + 0)^{-1} = 1. $$ Thus, the value of the limit is 1, which corresponds to option (D), none of the given options.

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