Problem 2.2.1 asks us to prove that several sequences converge. Here they are, together with a couple more:
a) $\displaystyle \lim \frac{1}{6n^2+1}=0$
b) $\displaystyle \lim \frac{3n+1}{2n+5}=\frac{3}{2}$
c) $\displaystyle \lim \frac{2}{\sqrt{n+3}}=0$
d) $\displaystyle \lim \frac{10n^2}{n^2-10}=10$
a) $\displaystyle \lim \frac{\cos(n)}{n^2+1}=0$
ETA: Went back on 09/16 and edited this response so it's not as gross looking and the answer's actually correct!
b.) $\lim \frac{3n+1}{2n+5} = \frac{3}{2}$
Let $\epsilon > 0$.
Then choose $N\in\mathbb{N} \ni N > \frac{13}{4\epsilon}$.
Then $n\geq N$ implies:
$$\bigg|\frac{3n+1}{2n+5} - \frac{3}{2}\bigg| = \frac{13}{4n+10} < \frac{13}{4n} \leq \frac{13}{4N} < \frac{13}{4\frac{13}{4\epsilon}} = \epsilon$$
Which implies $\bigg|\frac{3n+1}{2n+5} - \frac{3}{2}\bigg|<\epsilon$ so the sequence converges to $\frac{3}{2}$.$\blacksquare$
@qkhan Definitely a good start! We will learn how to improve it, though.
There's a template on page 41 for this sort of proof that you'd certainly want to follow. The first line should certainly be "Let $\varepsilon>0$" and the second line would specify exactly how to choose $N$. Note, though, that we expect a choice of the form $N>\text{something}$ but you've got $N<\text{something}$.
c) $\displaystyle \lim \frac{2}{\sqrt{n+3}}=0$
So for this one, I did as suggested and did some fooling around and I think I have an answer. It feels too simple though.
Claim:
c) $\displaystyle \lim \frac{2}{\sqrt{n+3}}=0$
Proof:
Let $\epsilon > 0$.
Choose $N \in \mathbb N$ such that $N > \frac{4}{\epsilon^2}$.
Then $ n \geq N$ implies:
$$\bigg|\frac{2}{\sqrt{n+3}}\bigg| = \frac{2}{\sqrt{n+3}} < \frac{2}{\sqrt{n}} \leq \frac{2}{\sqrt{N}} < \frac{2}{\sqrt{\frac{4}{\epsilon^2}}} = \frac{2}{\frac{2}{\epsilon}} = \epsilon $$
This all implies $\bigg|\frac{2}{\sqrt{n+3}}\bigg| < \epsilon$
I'm still at a loss as to how I should conclude this, but I would love some feedback (if I made simple math screwups don't be too mean it's like 1:30 in the morning).
Taking a stab at e (second a)
$\lim \dfrac{\cos(n)}{n^2+1} = 0$.
Let $\epsilon > 0$. Choose $N \in \mathbb{N}$ such that $N > \sqrt{\dfrac{1-\epsilon}{\epsilon}}$.
Thus, $N^2 > \dfrac{1}{\epsilon} - 1 \implies N^2 + 1 > \dfrac{1}{\epsilon}$. And so $\dfrac{1}{N^2+1} < \epsilon$.
Note $\cos(x) \leq 1, \forall \, x \in \mathbb{N}$.
Hence, $\dfrac{\cos(N)}{N^2+1} \leq \dfrac{1}{N^2+1} < \epsilon \implies \left\lvert \dfrac{\cos(N)}{N^2+1} - 0 \right\rvert < \epsilon$.
Therefore, the limit of the sequence is 0. $fin$
Okay, since no one has actually tried to do this one yet, I'm going to go ahead and try to take care of a.) (the first a, not the one that was already done).
a.) $\lim \frac{1}{6n^2+1} = 0.$
Proof:
Let $\epsilon$ > 0. Chose $N \in \mathbb{N}$ such that $N > \frac{1}{\sqrt{6\epsilon}}$. Thus by the definition of the limit, $ n \ge N \implies$
$$\left| \frac{1}{6n^2+1} - 0\right| = \frac{1}{6n^2+1} < \frac{1}{6n^2} \le \frac{1}{6N^2} < \frac{1}{6(\frac{1}{\sqrt{6\epsilon}})^2} = \frac{1}{\frac{1}{\epsilon}} = \epsilon $$
Therefore, $\lim\frac{1}{6n^2+1} = 0$ is true.
@violincounter Consider a simplification before solving for $N$. Sometimes, this is more than just a convenience - it's actually necessary. What is $N$, if $\varepsilon=2$, for example?
@jtamberi Why is $\epsilon$ less than 0 instead of greater than?
@ediazloa - Because I hit the wrong key on my keyboard at 8:30ish this morning and didn't realize I hit the wrong one.