# What is the limit of  f(x) = (2x+5)/abs(3x-4) as x goes to infinity?

Oct 18, 2015

$\frac{2}{3} \vee - \frac{2}{3}$, see the explanation.

#### Explanation:

$| 3 x - 4 | = 3 x - 4$ for $3 x - 4 \ge 0 , 3 x \ge 4 , x \ge \frac{4}{3}$

$| 3 x - 4 | = - 3 x + 4$ for $3 x - 4 < 0 , 3 x < 4 , x < \frac{4}{3}$

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

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

Note:
$\frac{5}{x} \to 0$ when $x \to \infty \vee x \to - \infty$
$\frac{4}{x} \to 0$ when $x \to \infty \vee x \to - \infty$