limit as $x$ approaches infinity of $\frac{1}{x}$ How can I show $\lim_{x \to \infty} \frac{1}{x} = 0$ using epsilon delta proof.
Its pretty obvious that the limit is zero, but I am still new at epsilon proofs.
 A: What you are trying to prove is that "if $x$ is large enough, then $\frac{1}{x}$ gets close enough to $0$."  Approaching infinity represents the concept of allowing $x$ to be arbitrarily large.
Here is how to state the proof, with some comments along the way.  Let $\varepsilon>0$ be some arbitrary real number (this will represent "error" or "distance from 0").  Let $N=\frac{1}{\varepsilon}$ (this represents how close to infinity $x$ must be).  Whenever $x>N$, then $\frac{1}{x}<\frac{1}{N}$, and since $\frac{1}{N}=\varepsilon$ we have $\left|\frac{1}{x}-0\right|<\varepsilon$ (that is, the distance between $\frac{1}{x}$ and $0$ is less than the error $\varepsilon$).  Therefore, $\lim_{x\to\infty}\frac{1}{x}=0$.  
A: Let $f:\Bbb R\setminus\{0\}\to\Bbb R$ be $f(x)=1/x$. You have conjectured that
$$
\lim_{x\to\infty}f(x)=0
$$
Rigorously, this means that for every $\varepsilon>0$ there exists an $M$ such that
$$
\left\lvert\frac{1}{x}\right\rvert<\varepsilon
$$
whenever $x>M$. Can you prove this?
A: For every $\epsilon > 0 $, there exists $N > 0$ such that if $ x > N $, then $|\frac{1}{x}|< \epsilon$.
Finding N,  we see,  $|\frac{1}{x}|< \epsilon \Rightarrow x > \frac{1}{\epsilon} \\$
Proof:
For $\epsilon > 0$ , take $N = \frac{1}{\epsilon}$ and  $x> N= \frac{1}{\epsilon} $
$\frac{1}{x} < \epsilon \Rightarrow |\frac{1}{x} |< \epsilon \\ $
Therefore, $\lim_{x\to\infty} \frac{1}{x} = 0 $
Hence, proved.
