$$\lim_{x\to 1} {\frac1x} = 1$$ Can anybody show me a complete proof using the definition. Thanks in advance!
By the definition means proving that: For every $\epsilon > 0$ there exist $\delta > 0$ where $|1/x - 1| < \epsilon$ when $0 < |x - 1| < \delta$.
On
Let $\epsilon > 0$. Pick $\delta = \min\left\{ 0.5 , \frac{\epsilon}{2} \right\}$.
Then, for any $x$ such that $|x - 1| < \delta$, we have $|x - 1| < 0.5$ and $|x - 1| < \frac{\epsilon}{2}$.
The first inequality tells us that $x = |x| > 0.5$. Multiplying the second inequality by 2 gives $$ 2|x-1| < \epsilon $$
Then,
$$ \left| 1 - \frac{1}{x} \right| = \frac{|x-1|}{|x|} < \frac{|x-1|}{0.5} = 2|x-1| < \epsilon $$
This proves that $\frac{1}{x}$ gets arbitrarily close to $1$ when $x \to 1$. We have shown that, for any given $\epsilon$, there is a choice of $\delta$ such that any $x$ value within $\delta$ of $x = 1$ will give us a value for $1/x$ that is within $\epsilon$ of 1.
EDIT: Here is my rationale for picking $\delta = \min\left\{ 0.5 , \frac{\epsilon}{2} \right\}$. I want to pick a $\delta$ such that eventually I get
$$ \left| 1 - \frac{1}{x} \right| < \epsilon$$
Rearranging this gives
$$ |x - 1 | < |x| \epsilon$$
Normally, if the $x$ weren't there, it would be very obvious what to choose for $\delta$ (namely it would just be whatever is on the right hand side of the inequality). $\delta$ can't depend on $x$ because logically that would not make sense (we have to pick $x$ based on $\delta$ later, so $\delta$ can't depend on $x$).
So, to get around the $x$ on the right hand side, we can simply choose to study a small region around $x = 1$. When you take limits, you only care about the stuff near the limit anyway, so this makes sense. Normally, I would choose $\delta < 1$ because 1 is a nice number, but $1/x$ blows up when we get to $x = 0$, so we have to refine a little further. $0.5$ is a nice number, but you can really just pick anything less than 1. You can pick just about any restriction for delta here, as long as it doesn't include any weird parts of the function you're taking a limit of.
In my case, I went with 0.5. So, by enforcing $\delta < 0.5$, I am guaranteeing that we only look at $x$ values from $0.5 < x < 1.5$. Generally speaking, the smaller a $\delta$ is, the more likely it is to work. On the right hand side, if we only consider $0.5 < x < 1.5$, the smallest this expression can be would be at $x = 0.5$. Thus, if we choose $\delta < 0.5 \epsilon$, we are guaranteed that $ |x - 1 | < |x| \epsilon$ will hold true.
We have placed two restrictions on $\delta$, so in order for both to be satisfied, $\delta$ must be the minimum of both of them.
On
$| \frac{1}{x} - 1 | = \frac{|1-x|}{|x|} = \frac{|x-1|}{|x-1 + 1|} \leq \frac{|x-1|}{|x-1| - 1}$
We need that $ \frac{|x-1|}{|x-1| - 1} < \epsilon$ for every $\epsilon>0$, now $|x-1|^2 < \epsilon |x-1| - \epsilon \Leftrightarrow |x-1|^2 - \epsilon|x-1| +\epsilon < 0$
by this way you find a $\delta(\epsilon)$ , such that $|x-1| < \delta$
If $|x-1| < \min(1/2,\epsilon/2)$, then $|x| > 1/2$ and
$$\left|\frac1{x}-1\right|= \frac{|x-1|}{|x|}< 2|x-1|<\epsilon$$