Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to solve $\displaystyle\lim_{x\to3^-}\frac{3}{x - 3} - \frac{3}{x} - 9$. Here's what I have tried.

$$ \lim_{x\to3^-}\frac{3}{x - 3} - \frac{3}{x} - 9 \\ \lim_{x\to3^-}\frac{3x}{x(x - 3)} - \frac{3(x-3)}{x(x-3)} - \frac{9x(x-3)}{x(x-3)} \\ \lim_{x\to3^-}\frac{3x - 3x - 9x^2 + 27x}{x(x - 3)} \\ \lim_{x\to3^-}\frac{- 9x^2 + 27x}{x^2 - 3x)} \\ $$

By L' Hopital's Rule,

$$ \lim_{x\to3^-}\frac{-18x + 27}{2x - 3} \\ \lim_{x\to3^-}\frac{-18}{2} \\ -9 \\ $$

However, Wolfram Alpha claims that the limit is infinity:


share|cite|improve this question
You've missed a +9 in the numerator of your third centred line. An easier approach would be to use limit of the sum equals sum of the limits, which applies in this case. – user12477 Sep 22 '12 at 21:27
Ah thanks, but $\lim_{x\to3^-}\frac{3}{x - 3}$ can't be calculated directly, right? – John Hoffman Sep 22 '12 at 21:28
Well, in a sense it can. The conventional answer is that the limit is $-\infty$, since the term $\frac{3}{x-3}$ is less than any given large negative number, provided $x$ is suffiently close to but less than 3. – user12477 Sep 22 '12 at 21:34
up vote 2 down vote accepted

In your second step, $3(x-3)=3x-9$, not $3x$. When you restore the missing term, you’ll find that l’Hospital’s rule no longer applies.

share|cite|improve this answer

You made an algebra mistake; instead of

$$\lim_{x\to3^-}\frac{3x - 3x - 9x^2 + 27x}{x(x - 3)}$$

It is

$$\lim_{x\to3^-}\frac{3x - 3x + 9 - 9x^2 + 27x}{x(x - 3)}$$

So you just forgot a term.

share|cite|improve this answer

You missed a term $+9$ on the numerator of your fraction. This matters because then the limit is not of the form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, which is a hypothesis of l'Hôpital's rule.

share|cite|improve this answer

The numerator and denominator of the fraction will both be divisible by $x-3$ if the limit exists. You'll have $$ \lim_{x\to3} \frac{(x-3)(\cdots\cdots\cdots)}{(x-3)(\cdots\cdots)}. $$ Then cancel the common factor. After that, you can plug in $3$ for $x$.

share|cite|improve this answer

Why not use WolframAlpha to help you find where the problem occurred? Your third line already evaluates to -9, so you know the error is not in the rest of your computation.

In fact the plot shows that your third line is identically equal to -9, which hints that you did something to incorrectly cancel out the $3/(x-3)$ and the $3/x$. I'm being deliberately vague here to highlight the fact that this basic level of information is in your grasp.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.