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Section 3.3: 

Motivation:  Graph $f(x) = {x^2 \over (x - 2)(x + 2)}$

%%Note $f'(x) = {2(x^2 + x + 4) \over (x^2 - 4)^2}$
%%and $f''(x) = {2(2x^3 + 3x^2 + 24x + 4) \over (x^2 - 4)^4}$

\vfill


\eject

To find vertical asymptotes,\hfil \break
 find all $a \in {\cal R}$ such that \hfil \break
$lim_{x \rightarrow a^-} f(x) = \pm \infty $ and/or
$lim_{x \rightarrow a^+} f(x) = \pm \infty $

Ex:  $f(x) = {1 \over (x+2)(x -3)^2}$

\vskip 1.1in

  Horizontal asymptotes/limits at infinity

\s

To find horizontal asymptotes: \hfil\break
 calculate $lim_{x \rightarrow +\infty} f(x)$ and  
$lim_{x \rightarrow -\infty} f(x)$ 

\s

IF $lim_{x \rightarrow +\infty} f(x) = L$ where $L$ is a finite real number, then $y = L$ is a 
horizontal asymptote.  
\s

IF $lim_{x \rightarrow -\infty} f(x) = K$ where $K$ is a finite real number, then $y = K$ is a 
horizontal asymptote.  

\vfill
\eject

Ex: $f(x) = {2x^3 - x^2 + 1 \over 8x^3 + x + 3}$ 

$lim_{x \rightarrow +\infty} {2x^3 - x^2 + 1 \over 8x^3 + x + 3} = $
\vfill

Similarly, $lim_{x \rightarrow -\infty} {2x^3 - x^2 + 1 \over 8x^3 + x + 3} = $

Horizontal asymptote(s):
\vskip 1cm
\eject

Ex: $f(x) = { x^2 + 1 \over 2x^5 + x^2 - 3}$   

$lim_{x \rightarrow +\infty} { x^2 + 1 \over 2x^5 + x^2 - 3} = $

\vfill

Similarly, $lim_{x \rightarrow -\infty} { x^2 + 1 \over 2x^5 + x^2 - 3} = $

Horizontal asymptote(s):
\v
\hrule

Ex: $f(x) = { 2x^5 + x^2 - 3 \over x^2 + 1 }$  

$lim_{x \rightarrow +\infty} { 2x^5 + x^2 - 3 \over x^2 + 1 } = $
\vfill

Also, $lim_{x \rightarrow -\infty} { 2x^5 + x^2 - 3 \over x^2 + 1 }= $

Horizontal asymptote(s):
\v
\eject

Ex: $f(x) = { 2x \over \sqrt {x^2 + 1} }  $  

$lim_{x \rightarrow +\infty} { 2x \over \sqrt{x^2 + 1} } = $
\vfill

$lim_{x \rightarrow -\infty} { 2x \over \sqrt{x^2 + 1} }= $
\vfill
\v\v\v\v\v\v\v\v\v\v\v
Horizontal asymptote(s):
\v
\eject

Ex:  $f(x) = x^2 - x^3$

$lim_{x \rightarrow +\infty} x^2 - x^3 = $
\vfill
\vfill\vfill
$lim_{x \rightarrow -\infty} x^2 - x^3 = $
\vfill\vfill
\vfill
Horizontal asymptote(s):
\v
\hrule

Ex:  $f(x) = x^{2 \over 3} - x$
 
$lim_{x \rightarrow +\infty} x^{2 \over 3} - x =$
\vfill
\vfill\vfill
$lim_{x \rightarrow -\infty} x^{2 \over 3} - x =$
\vfill
\vfill
\vfill
Horizontal asymptote(s):

\v

\end