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Question:Find the the limits of these sequence.if the limits does no exist ,explain why.

(1)$\left \{ cos((2n+1)\frac{\pi}{2}) \right \}_{n=1}^{\infty }$ my answer: so when n=1,lim=0

(2)$\left \{ \frac{\pi^{n}}{4^{n}} \right \}_{n=1}^{\infty }$ my answer:since n =1 , lim=$\pi/4$

(3)$\left \{ \frac{n^{2}+3}{n^{3}+n^{2}-1} \right \}_{n=1}^{\infty }$ my answer:when n=1,lim=4

(4)$\left \{ nsin(\frac{\pi}{n}) \right \}_{n=1}^{\infty }$ my answer:when n=1,lim=0

(5)$\left ( 1-\frac{1}{2} \right ),\left ( \frac{1}{2} -\frac{1}{3}\right ),\left ( \frac{1}{3}-\frac{1}{4} \right ),\left (\frac{1}{4} -\frac{1}{5}\right )\cdots $ I have no idea how to do this one

(6)$\left ( \sqrt{2}-\sqrt{3} \right ),\left ( \sqrt{3}-\sqrt{4} \right ),\left ( \sqrt{4}-\sqrt{5} \right )\cdots $

I have no idea how to do this one either

I want to know if my answers are correct for the question that I have answer and show me how to do the rest.I will be really grateful if i can get some help.

user136877
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    These are sequences, each tending to infinity, so you want to consider their behavior as $n \to \infty$, not when $n=1$. For example, the limit of the third sequence is zero. Notice that as $n$ gets very large, the denominator, a cubic, grows much faster than the numerator, a quadratic. This is the intuitive explanation. – solstafir Apr 16 '14 at 21:53
  • can you show me an example of (1) so I can relate to the other question? – user136877 Apr 16 '14 at 21:55
  • Okay, let $f(n)=\cos\frac{(2n+1)\pi}{2}$ for $n\in\mathbb{N}$. This is the sequence, and the question asks you to find the limit as $n\to\infty$. Because the cosine function is periodic, it will oscillate between two distinct limits, its max (1) and min (-1), as $n$ increases. This means that you can always find arbitrarily large values of $n$ for which the function is near each of these limits. Therefore, the limit does not exist. – solstafir Apr 16 '14 at 22:00
  • Also note that the first sequence is just {0,1,0,-1,...} i.e. the odd entries are zero and the even ones swap between 1 and -1. – mathematician Apr 16 '14 at 22:05
  • How about question (5) and (6), still dont know how to slove those two? – user136877 Apr 16 '14 at 22:19

2 Answers2

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Outline:

In the first case, $\cos((2n+1)\cdot \pi/2) = 0$, $\forall n \in \mathbb{Z^{+}}$. The limit is obvious. The next case is $0$ since $\pi < 4$ and the third is also $0$. Now, $\lim_{n \rightarrow \infty} n\cdot \sin(\pi n) = \lim_{m \rightarrow 0} \frac{\pi \cdot \sin(\pi m)}{\pi m} = \pi$. The next sequence goes as $1/[n\cdot (n+1)] \rightarrow 0$ and finally in the last case $\sqrt{n} - \sqrt{n+1} = \frac{-1}{\sqrt{n} + \sqrt{n+1}} \rightarrow 0$.

Solution to (3):

$$\lim_{n \rightarrow \infty} \frac{n^2 + 3}{n^3+n^2 -1} \\ = \lim_{n \rightarrow \infty} \frac{1/n + 3/n^3}{1 + 1/n - 1/n^3} \\ = \lim_{m \rightarrow 0} \frac{m + 3m^3}{1 + m - m^3} \\ = \frac{0 + 3\cdot 0^3}{1 + 0 - 0^3} = 0.$$

Note that the last step uses continuity at $0$; the second step uses convergence of the sequence given by $1/n$ to $0$.

Solution to (4):

$$\lim_{n \rightarrow \infty} n\cdot \sin(\pi/n) \\ = \lim_{m \rightarrow 0} \frac{\sin(\pi m)}{m} \\ = \lim_{m \rightarrow 0} \frac{\pi \cdot \sin(\pi m)}{\pi \cdot m} \\ = \pi \cdot \lim_{m \rightarrow 0} \frac{\sin(\pi m)}{\pi \cdot m} \\ = \pi \cdot \lim_{\pi m \rightarrow 0} \frac{\sin(\pi m)}{\pi \cdot m} = \pi.$$

The last limit is known to be $1$, which is made rigorous by l'Hopitals Rule where $\cos(0)/1 = 1$. Note that convergence is implicitly used here as well; I will spare the analysis for sake of simplicity.

Christopher K
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  • I dont quite understand how you did with the question (3) and (4). in question 3, the denominator is geeting bigger,is this the reason the limit is 0? – user136877 Apr 16 '14 at 22:26
  • Yes, it is somewhat intuitive, but see above for a step-by-step "proof". – Christopher K Apr 16 '14 at 22:44
  • thank you!!!! I still have a continuity problem unsolved, can u plz look at it if u have time?http://math.stackexchange.com/questions/756855/intermediate-value-theorem-problem – user136877 Apr 16 '14 at 22:45
  • @user136877, I added that in the form of a comment on the other page. – Christopher K Apr 16 '14 at 22:52
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(5) For the sequence $(\frac{1}{n}-\frac{1}{n+1})_{n=1}^{\infty}$, note that $\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n(n+1)}$.

David H
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