# What is the limit of (2x+3)/(5x+7) as x goes to infinity?

##### 2 Answers
Oct 4, 2015

$\lim \left(x \to \infty\right) \frac{2 x + 3}{5 x + 7} = \frac{2}{5}$

#### Explanation:

$\lim \left(x \to \infty\right) \frac{2 x + 3}{5 x + 7}$

Notice that the degree of the numerator and the denominator are

the same i.e: 1, for this and all the similar scenarios the limits is

simply the ratio of the leading coefficients of top to bottom:

$\therefore \lim \left(x \to \infty\right) \frac{2 x + 3}{5 x + 7} = \frac{2}{5}$

Oct 4, 2015

As $x \to \infty$, the $3$ and $7$ become insignificant relative to the magnitude of $x$. In other words, $x$ becomes so large that:

${\lim}_{x \to \infty} \frac{2 x + 3}{5 x + 7} = {\lim}_{x \to \infty} \frac{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{5 \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}$

$= {\lim}_{x \to \infty} \frac{2}{5}$

$= \textcolor{b l u e}{\frac{2}{5}}$