# What is the horizontal asymptote of y=((2x+3)(5x-2))/(7x^2-3) ?

Sep 11, 2014

Its horizontal asymptote is $y = \frac{10}{7}$.

By taking the limits at infinity,
${\lim}_{x \to \infty} \frac{\left(2 x + 3\right) \left(5 x - 2\right)}{7 {x}^{2} - 3}$
by divide the numerator and the denominator by ${x}^{2}$,
$= {\lim}_{x \to + \infty} \frac{\left(2 + \frac{3}{x}\right) \left(5 - \frac{2}{x}\right)}{7 - \frac{3}{x} ^ 2} = \frac{\left(2 + 0\right) \left(5 - 0\right)}{7 - 0} = \frac{10}{7}$

Similarly, you can find
${\lim}_{x \to - \infty} \frac{\left(2 x + 3\right) \left(5 x - 2\right)}{7 {x}^{2} - 3} = \frac{10}{7}$

Hence, there is only one horizontal asymptote $y = \frac{10}{7}$.