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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: http://www.wolframalpha.com/input/?i=lim+x+-%3E+3+%5Cfrac%7B3%7D%7Bx+-+3%7D+-+%5Cfrac%7B3%7D%7Bx%7D+-+9

Why?

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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
1  
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

5 Answers 5

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.

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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.

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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.

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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 $x$.

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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.

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