I'm trying to solve $$ \lim_{x\to\infty}\frac{3x+5}{x-4} $$
Since the numerator and denominator both increase without bound, I try to get something more useful by dividing everything by $ x $.
$$
\begin{align}
\lim_{x\to\infty} \frac{3x+5}{x-4}
= \lim_{x\to\infty} \frac{\frac{3x+5}{x}}{\frac{x-4}{x}}
= \lim_{x\to\infty} \frac{3 + \frac{5}{x}}{ \frac{-4}{x} }
= \frac{ \lim_{x\to\infty} 3 + \frac{5}{x} }{ \lim_{x\to\infty} \frac{-4}{x}}
\end{align}
$$
This gets me a numerator approaching 3 and a denominator approaching zero. But since the quotient law for limits specifically excludes a zero denominator, I don't know what do from here.
Wolfram Alpha says the limit is 3 (and a graph agrees) so I think I'm on the right track, but I have a feeling I messed up the algebra somewhere.
Thanks.
Best Answer
$\begin{eqnarray*}\frac{x-4}{x}&=&\frac{x}{x}-\frac{4}{x} \\ &=& 1 - \frac{4}{x} \end{eqnarray*}$