so I'm taking calc 2 right now and so far the proofs/applying theorems to every question is kicking my ass since I'm so used to math being so computational. Anyway, one of the questions I have is the one above. I ended up just rehashing the definition of the limit and trying to replace a(n) with given sequence and the L with the given L. I don't really understand it.
I ended up writing my answer as lim(n->infinity) 5 + lim(n->infinity) 1/n = 0 and then making a(n) = 1/n There is an N s.t. if n > N, then |1/n - 0| < epsilon - a(n) and you can see where this is going and it's obviously wrong..I understand the definition of a limit on its own, and can envision it on the graph, but don't know exactly how I can actually apply it to a given sequence and its limit.
Let $a_n=5+\frac 1n$.
The way you think about it is that you want to find $N$ such that $|a_n-5|<\epsilon$, i.e. that $|\frac 1n|<\epsilon$ whenever $n>N$.
All you need to do is rearrange this inequality to solve for $n$, giving $n>\frac 1\epsilon$. But of course, the $N$ you choose needs to be an integer, so you simply round it up to the nearest integer, giving $N=\lceil \frac 1\epsilon \rceil $.
The above should be done either in your head or on a piece of scrap paper. Once you obtain this value of $N$, you present your proof by writing it out from the definition:
$$\forall \epsilon>0 \; \exists \; N=\lceil \frac 1\epsilon \rceil \in \Bbb N \; | \; n>N \implies |a_n-5|=|\frac 1n|=\frac 1n<\frac 1N=\frac{1}{\lceil \frac 1\epsilon \rceil}≤\frac{1}{\frac 1\epsilon}=\epsilon$$
This implies that $$\lim_{n\rightarrow \infty}a_n=5$$