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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 $$H \cdot A$$

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