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?

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 
$$
\lim_{x \to 2}\frac{1}{1-x} = -1.
$$
