<s>[INST] Why do we find vertical asymptote of functions (in rational fractions) by turning the denominator into $0$? Like for $$f(x) = \frac{1}{x-2}$$ we can find its vertical asymptote by finding what $x$ can be in order to turn the denominator into zero. 
e.g. $x=2$
But why is that? Why does the vertical asymptote correspond to the zero of the denominator?
 [/INST] A function $f$ has a vertical asymptote at $x = a$ if $|f(x)|$ increases without bound as $x$ approaches $a$, that is, $f$ has a vertical asymptote if one or more of the following conditions holds:
\begin{align*}
\lim_{x \to a^+} f(x) & = \infty\\
\lim_{x \to a^+} f(x) & = -\infty\\
\lim_{x \to a^-} f(x) & = \infty\\
\lim_{x \to a^-} f(x) & = -\infty
\end{align*}
Why does the function 
$$f(x) = \frac{1}{x - 2}$$
have a vertical asymptote at $x = 2$?  
Consider what happens as $x$ approaches $2$ from the right.
$$
\begin{array}{c c c}
x & x - 2 & f(x) = \dfrac{1}{x - 2}\\ \hline
2.1 & 0.1 & 10\\
2.01 & 0.01 & 100\\
2.001 & 0.001 & 1000\\
2.0001 & 0.0001 & 10000
\end{array}
$$
Observe that as $x$ approaches $2$ from the right, we can make $f(x)$ as large as we want by making $x - 2$ sufficiently close to $0$.  Consequently, $f$ increases without bound as $x$ approaches $2$ from the right.  We write
$$\lim_{x \to 2+} f(x) = \infty$$
read the limit as $x$ approaches $2$ from above (or from the right) is infinity, to indicate that the function increases without bound as $x$ approaches $2$ from the right.  Since this condition holds, the function $f(x) = \frac{1}{x - 2}$ has a vertical asymptote at $x = 2$.
Consider what happens as $x$ approaches $2$ from the left.
$$
\begin{array}{c c c}
x & x - 2 & f(x) = \dfrac{1}{x - 2}\\ \hline
1.9 & -0.1 & -10\\
1.99 & -0.01 & -100\\
1.999 & -0.001 & -1000\\
1.9999 & -0.0001 & -10000
\end{array}
$$
Observe that $x$ approaches $2$ from the left, we can make $f(x)$ as large a negative number as we like by taking $x - 2$ sufficiently close to $0$.  Consequently, $f$ decreases without bound as $x$ approaches $2$ from the left.  We write 
$$\lim_{x \to 2-} f(x) = -\infty$$
read the limit of $f(x)$ as $x$ approaches $2$ from below (or from the left) is negative infinity, to indicate that $f$ decreases without bound as $x$ approaches $2$ from the left.  This condition also implies that the function $f(x) = \frac{1}{x - 2}$ has a vertical asymptote at $x = 2$.  

The reason why the function $f(x) = \frac{1}{x - 2}$ has a vertical asymptote at $x = 2$ is that as $x$ approaches from the right or from the left, the denominator approaches $0$ while the numerator does not.  This occurs near the zero of the denominator since we can make the denominator arbitrarily small without making the numerator arbitrarily small.
More generally, for a rational function in simplest form, the function has a vertical asymptote at points where the denominator is equal to zero and the numerator is not equal to zero.  It is not sufficient that the denominator equal zero.
Consider the function 
$$g(x) = \frac{x^2 - 4}{x - 2}$$
Notice that 
\begin{align*}
g(x) & = \frac{x^2 - 4}{x - 2}\\
     & = \frac{(x + 2)(x - 2)}{x - 2}\\
     & = x + 2 && \text{provided that $x \neq 2$}
\end{align*}
While both $f(x) = \frac{1}{x - 2}$ and $g(x) = \frac{x^2 - 4}{x - 2}$ are undefined at $x = 2$ since the denominator is equal to $0$ when $x = 2$, the function $g$ does not have a vertical asymptote at $x = 2$.  As $x$ approaches $2$, $g(x)$ approaches $2 + 2 = 4$, so $|g(x)|$ does not increase without bound as $x$ approaches $2$.  Instead, the graph of $g$ is a punctured line.
 
</s>