# How do you find vertical, horizontal and oblique asymptotes for y=1/(2-x)?

May 31, 2016

Vertical asymptote $x = 2$
Horizontal asymptote $y = 0$

#### Explanation:

$\textcolor{b l u e}{\text{Determine vertical asymptotes}}$

Mathematically you are not allowed to divide by 0. This situation is called 'undefined'.

So at $x = 2$ the equation is undefined. Thus $x = 2$ is an asymptote.

Suppose $x > 2$ then $2 - x < 0$

As $x$ becomes increasingly closer to 2 then $2 - x$ becomes smaller and smaller but still negative.

In the same way, when $x < 2$ but approaching 2 then $2 - x$ becomes smaller and smaller but is still positive.
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So $\frac{1}{2 - x}$ becomes increasingly larger but negative or positive depending on what side of 2 we find $x$ to be.

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When writing limits; when approaching a value from the positive side then for the value 2 it would be written as ${\textcolor{w h i t e}{}}^{+} 2$

Same way for approaching from the negative side it is ${\textcolor{w h i t e}{}}^{-} 2$

Thus

$y = {\lim}_{x \to {\textcolor{w h i t e}{}}^{+} 2} \frac{1}{2 - x} = - \infty \text{ } \leftarrow \left(x > 2\right)$

$y = {\lim}_{x \to {\textcolor{w h i t e}{}}^{-} 2} \frac{1}{2 - x} = + \infty \text{ } \leftarrow \left(x < 2\right)$

$\textcolor{g r e e n}{\text{So "x=2 " is the vertical asymptote}}$

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$\textcolor{b l u e}{\text{Determine horizontal asymptotes}}$

$y = {\lim}_{x \to {\textcolor{w h i t e}{}}^{+} \infty} \frac{1}{2 - x} = \frac{1}{2 - \infty} = - \frac{1}{\infty} = {\textcolor{w h i t e}{}}^{-} 0$

$y = {\lim}_{x \to {\textcolor{w h i t e}{}}^{-} \infty} \frac{1}{2 - x} = \frac{1}{2 + \infty} = + \frac{1}{\infty} = {\textcolor{w h i t e}{}}^{+} 0$

$\textcolor{g r e e n}{\text{So "y=0 " is the horizontal asymptote}}$ 