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• Notes

Find the vertical and horizontal asymptotes for the graph of $f$ if they exist:

$f(x)=\frac{x-4}{|x|-4}$

$V, A \rightarrow|x|-4=0 \rightarrow|x|=4 \rightarrow x=\pm 4$

$x=4$

$\lim _{x \rightarrow 4} \frac{x-4}{|x|-4}=\lim _{x \rightarrow 4} \frac{x-4}{x-4}=1 \quad x=4$ is not V. $A$

$x=-4$

$\lim _{x \rightarrow -4} \frac{x-4}{|x|-4}=\lim _{x \rightarrow -4} \frac{x-4}{-x-4}=\lim \frac{-4-4}{-(-4)-4}=\frac{-8}{0}=\pm \infty$

$\lim _{x \rightarrow-4^{+}} \frac{x-4}{-x-4}=\frac{-8}{0^{-}}=+\infty$

$\lim _{x \rightarrow {-4}^{-}} \frac{x-4}{-x-4}=-\infty$

$x=4$ is $V \cdot A$

$H \cdot A \rightarrow \lim _{x \rightarrow \infty} \frac{x-4}{|x|-4}$

$|x|=x$ if $x \rightarrow \infty$

$\lim _{x \rightarrow \infty} \frac{x-4}{x -4}=1$

$\lim _{x \rightarrow-\infty} \frac{x-4}{|x|-4}$

$|x|=-x \quad$ if $x \rightarrow-\infty$

$\lim _{x \rightarrow-\infty} \frac{x-4}{-x-4}=\lim _{x \rightarrow-\infty} \frac{x(1-4/ x)}{x(-1-{4}/ {x})}$

$=\lim _{x \rightarrow-\infty} \frac{1-4 / x}{-1-4 / x}=\lim _{x \rightarrow -\infty} \frac{1}{-1}=-1$

$y=\pm 1$ are $H \cdot A$