009A Sample Final A, Problem 6

6. Find the vertical and horizontal asymptotes of the function  $f(x)={\frac {\sqrt {4x^{2}+3}}{10x-20}}.$

Foundations:
Vertical asymptotes occur whenever the denominator of a rational function goes to zero, and  it doesn't cancel from the numerator.
On the other hand, horizontal asymptotes represent the limit as x goes to either positive or negative infinity.

 Solution:

Vertical Asymptotes:
Setting the denominator to zero, we have
$0=10x-20=10(x-2),$
which has a root at $x=2.$ This is our vertical asymptote.

Horizontal Asymptotes:
More work is required here. Since we need to find the limits at $\pm \infty$, we can multiply our $f(x)$ by
${\frac {\sqrt {\frac {1}{x^{2}}}}{\,\,\,{\frac {1}{x}}}}.$
This expression is equal to 1 for positive values of x, and is equal to -1 for negative values of x. Since multiplying f(x) by an expression equal to 1 doesn't change the limit, we will add a negative sign to it when considering the limit as x goes to $-\infty$. Thus,
$\lim _{x\rightarrow \pm \infty }{\frac {\sqrt {4x^{2}+3}}{10x-20}}\,\,\cdot \,\,\pm {\frac {\sqrt {\frac {1}{x^{2}}}}{\,\,\,{\frac {1}{x}}}}=\lim _{x\rightarrow \pm \infty }\pm {\frac {\sqrt {{\frac {4x^{2}}{x^{2}}}+{\frac {3}{x^{2}}}}}{{\frac {10x}{x}}-{\frac {20}{x}}}}=\lim _{x\rightarrow \pm \infty }\pm {\frac {\sqrt {4+{\frac {3}{x^{2}}}}}{10-{\frac {20}{x}}}}=\pm {\frac {2}{10}}=\pm {\frac {1}{5}}$
Thus, we have a horizontal asymptote at y = -1/5 on the left (as x goes to $-\infty$), and a horizontal asymptote at y = 1/5 as x goes to $+\infty$).

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