How do you find the horizontal asymptote for y=(13x)/(x+34)?

Dec 7, 2015

Find ${\lim}_{x \to \pm \infty} \frac{13 x}{x + 34} = 13$ hence asymptote $y = 13$

Explanation:

One way is to divide both numerator and denominator by $x$ and look at the limit as $x \to \pm \infty$

$f \left(x\right) = \frac{13 x}{x + 34} = \frac{13}{1 + \frac{34}{x}}$

So ${\lim}_{x \to \infty} f \left(x\right) = \frac{13}{1 + 0} = 13$

and ${\lim}_{x \to - \infty} f \left(x\right) = \frac{13}{1 + 0} = 13$

So the horizontal asymptote is $y = 13$

In general if $f \left(x\right) = \frac{p \left(x\right)}{q \left(x\right)}$ for some polynomials $p \left(x\right)$ and $q \left(x\right)$ of the same degree, then $f \left(x\right)$ will have a horizontal asymptote equal to the quotient of the leading coefficients.

For example:

${\lim}_{x \to \infty} \frac{3 {x}^{2} + 4 x - 5}{2 {x}^{2} - 11 x + 1} = {\lim}_{x \to \infty} \frac{3 + \frac{4}{x} - \frac{5}{{x}^{2}}}{2 - \frac{11}{x} + \frac{1}{{x}^{2}}} = \frac{3}{2}$