Limit of a rational function where denominator approaches zero

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.

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You dropped a term from your denominator in the third step; it should be $1 - \frac{4}{x}$. – Qiaochu Yuan Jul 30 '11 at 23:45
Right you are! I suspected it was a dumb algebra mistake. – friedo Jul 30 '11 at 23:58

1 Answer

$\begin{eqnarray*}\frac{x-4}{x}&=&\frac{x}{x}-\frac{4}{x} \\ &=& 1 - \frac{4}{x} \end{eqnarray*}$

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Thanks -- my suspicion of an algebra mistake was correct. Now I get 3/1 which makes everything work. – friedo Jul 30 '11 at 23:57
@DLim: Thank you for the editing in the extra step to clarify. (Really, no reason to apologize.) – Jonas Meyer Aug 1 '11 at 14:52