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Chapter 1: Problem 83

The function(s) which have a limit as \(\mathrm{n} \rightarrow \infty\) (A) \(\left(\frac{n-1}{n+1}\right)^{2}\) (B) \((-1)^{n}\left(\frac{n-1}{n+1}\right)^{2}\) (C) \(\frac{n^{2}+1}{n}\) (D) \((-1)^{n} \frac{n^{2}+1}{n}\)

Short Answer

Expert verified
(A) \(\left(\frac{n-1}{n+1}\right)^{2}\) (B) \((-1)^{n}\left(\frac{n-1}{n+1}\right)^{2}\) (C) \(\frac{n^{2}+1}{n}\) (D) \((-1)^{n} \frac{n^2+1}{n}\) Answer: (A)

Step by step solution

01

Analyze function (A)

For the function (A) \(\left(\frac{n-1}{n+1}\right)^{2}\), as \(n \rightarrow \infty\), both numerator and denominator will also approach infinity. So, we can rewrite the function as: \[ \lim_{n\to\infty} \left(\frac{n-1}{n+1}\right)^{2} = \lim_{n\to\infty} \left(\frac{n}{n}\right)^{2} = \lim_{n\to\infty} 1^{2} = 1 \] Therefore, the limit exists for function (A).
02

Analyze function (B)

For the function (B) \((-1)^{n}\left(\frac{n-1}{n+1}\right)^{2}\), if we can find the limit of the expression without the \((-1)^n\) term, then we can decide the limit of the whole function. We know from Step 1 that the limit of \(\left(\frac{n-1}{n+1}\right)^{2}\) as \(n \rightarrow \infty\) is 1. However, the \((-1)^n\) term oscillates between -1 and 1 as n increases. Since the product of these two expressions cannot converge to a single value, we conclude that the limit does not exist for function (B).
03

Analyze function (C)

For the function (C) \(\frac{n^{2}+1}{n}\), we can use the dominant term method as \(n \rightarrow \infty\). The dominant term of the numerator is \(n^2\), and the dominant term of the denominator is \(n\). If we divide both numerator and denominator by \(n^2\), we get: \[ \lim_{n\to\infty} \frac{\frac{n^2}{n^2}+\frac{1}{n^2}}{\frac{n}{n^2}} = \lim_{n\to\infty} \frac{1+\frac{1}{n^2}}{\frac{1}{n}} =\lim_{n\to\infty} n (1+\frac{1}{n^2}) = \infty \] So, the function (C) does not have a limit as \(n \rightarrow \infty\).
04

Analyze function (D)

For the function (D) \((-1)^{n} \frac{n^2+1}{n}\), we already know from Step 3 that the limit of \(\frac{n^2+1}{n}\) does not exist as \(n \rightarrow \infty\). The oscillation of the \((-1)^n\) term does not affect the limit of the whole function. Therefore, function (D) also does not have a limit as \(n \rightarrow \infty\). In conclusion, only function (A) has a limit as \(n \rightarrow \infty\).

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Most popular questions from this chapter

The value of $\lim _{n \rightarrow \infty}\left(\frac{n !}{n^{n}}\right)^{\frac{2 n^{\prime}+1}{5 n^{2}+1}}$ is equal to (A) \(]\) (B) 0 (C) \(\left(\frac{1}{\mathrm{e}}\right)^{2 / 5}\) (D) \(\mathrm{e}^{2 / 3}\)

Consider the function \(f(x)=\left(\frac{a x+1}{b x+2}\right)^{x}\) where \(a^{2}+b^{2} \neq 0\) then \(\lim f(x)\) (A) exists for all values of \(a\) and \(b\) (B) is zero for \(\mathrm{a}<\mathrm{b}\) (C) is non existent for \(\mathrm{a}>\mathrm{b}\) (D) is e \({ }^{-\left(\frac{1}{a}\right)}\) or \(e^{-\left(\frac{1}{b}\right)}\) if \(a=b\)

Which of the following functions has two horizontal asymptotes (A) \(\mathrm{y}=\frac{|\mathrm{x}|}{\mathrm{x}+1}\) (B) \(\mathrm{y}=\frac{2 \mathrm{x}}{\sqrt{\mathrm{x}^{2}+1}}\) (C) \(y=\frac{\sin x}{x^{2}+1}\) (D) \(y=\cot ^{-1}(2 x+1)\)

\(\mathrm{A}_{0}\) is an equilateral triangle of unit area, \(\mathrm{A}_{0}\) is divided into four equal parts, each an equilateral triangle, by joining the mid points of the sides of \(\mathrm{A}_{0}\). The central triangle is removed. Treating the remaining three triangles in the same way of division as was done to \(\mathrm{A}_{0}\), and this process is repeated \(\mathrm{n}\) times. The sum of the area of the triangles removed in \(\mathrm{S}_{\mathrm{n}}\) then $\lim _{\mathrm{n} \rightarrow \infty} \mathrm{S}_{\mathrm{n}}$ is (A) \(1 / 2\) (B) 1 (C) \(-1\) (D) 2

For which of the following functions, Approx \(\mathrm{f}(\mathrm{x})\) exists : (A) \(\underset{x \rightarrow 1}{\text { Approx }} \frac{x^{2}-1}{|x-1|}\) (B) Approx \(\frac{2\\{x\\}-4}{[x]-3}\) (C) $\underset{x \rightarrow 0}{\operatorname{Approx}} \frac{1}{2-2^{\frac{1}{x}}}$ (D) None of these
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