How can one show that limit of $\frac{1}{1-x}$ as $x$ goes to $2$ exists? How can one show that limit of $\frac{1}{1-x}$ as $x$ goes to $2$ exists?
Its limit value is $-1$.
How can I prove this using epsilon and delta?
 A: Initially we may guess that 
$$
\frac{1}{1-x} \to -1
$$
as $x \to 2$. To prove this, note that we have $x\neq 1$ only if
$$
\bigg| \frac{1}{1-x} - (-1) \bigg| = \bigg|\frac{2-x}{1-x} \bigg| = \bigg|\frac{x-2}{x-1} \bigg|;
$$
we have $0 < |x-2| < 1/2$ only if $||x-1|-1| \leq |x-2| < 1/2$, only if $1/2 < |x-1|$, and only if
$$
\bigg| \frac{x-2}{x-1} \bigg| < 2|x-2|;
$$
given any $\varepsilon > 0$, we have $0 < |x-2| < \varepsilon/2$ only if $2|x-2| < \varepsilon$; hence
we have proved this:
for every $\varepsilon > 0$, we have $0 < |x-2| < \min \{1/2, \varepsilon/2 \}$ only if
$$
\bigg| \frac{1}{1-x} + 1 \bigg| < \varepsilon,
$$
which shows that 
$$
\lim_{x \to 2}\frac{1}{1-x} = -1.
$$
A: We first choose $\delta'= 1/2$.  Then, for $0<|x-2|<\delta'$, we have $\frac{1}{x-1}<2$.  Therefore, for all $\epsilon >0$, we have
$$\begin{align}
\left|\frac{1}{1-x}+1\right|&=\left|\frac{x-2}{1-x}\right|\\\\
&<2|x-2|\\\\
&<\epsilon
\end{align}$$
whenever $|x-2|<\delta=\min\left(\frac{\epsilon}{2},\frac12\right)$.  And we are done!
A: Define $f(x) = \frac{1}{1-x}$. We want to show that $f(x)$ is continuous at $-1$. With $\epsilon-\delta$ we need to show that for any $\epsilon>0$ there exists $\delta>0$ such that $|x-2|<\delta$ yields $|f(x)+1|<\epsilon$. This much is definition. Now
$$f(x)+1 = \frac{1}{1-x}+1 = \frac{x-2}{x-1}$$
I claim that for a given $\epsilon$, it is enough to choose $\delta = \frac{\epsilon}{1+\epsilon}$. Let's check if it works:
Note that if $|x-2|<\delta$, then $-\delta<x-2<\delta$, so $-\delta+1<x-1<\delta+1$. But then my $\delta<1$, so
$$
\frac{1}{1+\delta}<\frac{1}{x-1}<\frac{1}{1-\delta}\Longrightarrow \frac{1}{|x-1|}<\frac{1}{1-\delta}
$$
So we have
$$
|f(x)+1| = \left|\frac{x-2}{x-1}\right|<\frac{\delta}{1-\delta}=\frac{\frac{\epsilon}{1+\epsilon}}{1-\frac{\epsilon}{1+\epsilon}}=\epsilon
$$
Done! Now since the function is continuous at $-1$, its limit is its value. (I intentionally took a longer approach to try and familiarize the $\epsilon-\delta$ approach as much as I can)

If you are wondering how in the world I came up with this $\delta$, here's how you reverse engineer it (I have to repeat parts of previous argument, sorry!): We think in reverse direction, suppose we already know that $\delta$ exists, then what can it be? Whatever it is we should only care about small values of $\delta$ since limit only cares about small neighborhoods. So let's assume our unknown $\delta$ is less than one. Then if $\delta$ exists (as we are assuming) we must have (similar to above)
$$
\left|\frac{x-2}{x-1}\right|<\frac{\delta}{1-\delta}
$$
So it only remains to choose $\delta$ so that the right hand side is actually $\epsilon$ or less. Well solve $\frac{\delta}{1-\delta}=\epsilon$ and find $\delta=\frac{\epsilon}{1+\epsilon}$.
A: Where $I$ is a sufficiently close interval to the limit point, a limit will exists if:
Right Sided $\forall\varepsilon > 0\;\exists \delta >0 \;\forall x \in I \;(0 < x -2 < \delta \Rightarrow \lvert \frac 1{1-x} +1 \rvert<\varepsilon)$
Left Sided $\forall\varepsilon > 0\;\exists \delta >0 \;\forall x \in I \;(0 < 2-x < \delta \Rightarrow \lvert \frac 1{1-x} +1 \rvert<\varepsilon)$
So pick an arbitrary small number $\varepsilon$ and show the existence of some $\delta$ so that when $x$ lies within the restriction, then $\lvert \frac 1{1-x} +1 \rvert<\varepsilon$.
Or in simpler terms: that as the argument approaches the limit point the function converges towards the limit value.
A: Hint: Do the algebra and use the inequality that $|1 - x| > 1/2$ for $x > 3/2$:
$$\left|\frac{1}{1 - x} - (-1)\right| = \left|\frac{2 - x}{1 - x}\right| < 2|x - 2| < \cdots.$$
A: @Garsa : The $\varepsilon-\delta$ argument cannot be used to evaluate a limit; instead, it can only be used to prove that a conjectured value $l$ is the limit of a function at a given point.
In order to guess a possible value for the limit $\displaystyle \lim_{x\to 2} \frac{1}{1-x}$, one possible way is to plug $x=2$ into the function itself (assuming the function is definite in $2$): this way you get the guess:
$$l=-1\; .$$
In order to prove that $\displaystyle \lim_{x\to 2} \frac{1}{1-x} = -1$, you should prove that every time you choose an $\varepsilon >0$ "small", you can find a $\delta >0$ "small" in such a way that $|x-2|<\delta$ implies $\left| \frac{1}{1-x} - (-1)\right|<\varepsilon$. In practice, this means that once you fix a "small" value of the parameter $\varepsilon >0$, you can always cut out from the set of solutions of the parametric inequality $\left| \frac{1}{1-x} - (-1)\right|<\varepsilon$ a "small" symmetric neighbourhood of $2$.
Now the inequality:
$$\left| \frac{1}{1-x} - (-1)\right|<\varepsilon$$
is equivalent to the system:
$$\begin{cases} \frac{2-x}{1-x} < \varepsilon\\ \frac{2-x}{1-x}> - \varepsilon\end{cases}$$
i.e.:
$$\begin{cases} \frac{(2-\varepsilon)- (1-\varepsilon) x}{1-x} <0\\ \frac{(2+\varepsilon) - (1+\varepsilon) x}{1-x} >0\end{cases}$$
The latter system can be solved explicitly: in fact, assuming $0<\varepsilon <1$ (you don't lose generality, because $\varepsilon$ has to be "small"), you find the solutions:
$$\frac{2+\varepsilon}{1+\varepsilon} < x < \frac{2-\varepsilon}{1-\varepsilon}\; .$$
Finally, you have to cut a symmetric neighbourhood of $2$ out of these solutions. In order to do this, subtract $2$ in each side and find:
$$-\frac{\varepsilon}{1+\varepsilon} < x-2 < \frac{\varepsilon}{1-\varepsilon}\; ,$$
hence, if you let:
$$\delta := \min \left\{ \frac{\varepsilon}{1+\varepsilon} , \frac{\varepsilon}{1-\varepsilon}\right\} = \frac{\varepsilon}{1+\varepsilon}$$
it is obvious that the neighbourhood $-\delta <x-2<\delta$ is contained in the set $\frac{2+\varepsilon}{1+\varepsilon} < x < \frac{2-\varepsilon}{1-\varepsilon}$.
Therefore, if $|x-2|<\delta$ then $|f(x) - (-1)| < \varepsilon$ and thus $\displaystyle \lim_{x\to 2} f(x) = -1$.
A: The definition of convergence says a sequence $\{a_n\}$ converges to a limit $L$ if given an $\epsilon$ however small there exists $N(\epsilon)$ such that $|a_n-L|<\epsilon$ for all $n>N(\epsilon)$.
So let $x_n$ be values getting progressively closer to 2. I.e. For any $\epsilon_2$ there exists $N_2(\epsilon_2)$ such that $|x_n-2|<\epsilon$ for all $n>N_2(\epsilon_2)$.
By triangle inequality $|x_n-1|<|x_n-2|+1=\epsilon_2+1$
Let us show that given $\epsilon$ there does exist $N(\epsilon)$ such that $|\frac{1}{1-x_n}--1|<\epsilon$ for all $n>N$.
$$|\frac{1}{1-x_n}--1|<\epsilon$$
$$\Leftarrow |\frac{2-x_n}{1-x_n}|<\epsilon$$
$$\Leftarrow \frac{|2-x_n|}{|1-x_n|}<\epsilon$$
$$\Leftarrow \frac{\epsilon_2}{\epsilon_2+1}<\epsilon$$
$$\Leftarrow \epsilon_2\epsilon<\epsilon_2+1$$
$$\Leftarrow 1<\epsilon_2(1-\epsilon)$$
$$\Leftarrow \frac{1}{1-\epsilon}<\epsilon_2$$
So given any $\epsilon$ choose $\epsilon_2>\frac{1}{1-\epsilon}$ and we find a find a corresponding value $x_n$ (as we defined $x_n$ as converging) such that 
$|\frac{1}{1-x_n}|<\epsilon$.
A: Let $f(x)=1$ for every $x\in \mathbb {R}$ and $g(x)=1-x$ for every $x\in \mathbb {R}$. Notice that $\lim_{x\rightarrow 2}f(x)=1$; to see the $\epsilon-\delta$ argument for this notice that for this function $f(x)$, choose $\delta =\epsilon$ so that we have, $|f(x)-1|=0\leq |x-2|\leq \delta=\epsilon.$
Further observe that $\lim_{x\rightarrow 2}g(x)=-1$; To see this, choose $\delta =\epsilon$ again so that we have, $|g(x)-(-1)|=|1-x+1|=|2-x|=|x-2|\leq \delta =\epsilon.$   
So both the limit exists and $\lim_{x\rightarrow 2}g(x)$ is non-zero finite number. Hence the algebra of limits guarantees that 
$$\lim_{x\rightarrow 2}\frac{f(x)}{g(x)}=\frac{1}{-1}=-1$$.
