Why this limit can't be computed like this?

This is the limit: $$\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$$

The solution manual says it is: $\frac{11}{4}$

I've tried to solve it like a polynomial like this:

$$\frac{(\frac{3x^2}{3x^2}-\frac{x}{3x^2}-\frac{10}{3x^2})*3x^2}{(\frac{x^2}{x^2}-\frac{4}{x^2})*x^2}=$$

$$= \frac{(1-0-0)*3x^2}{(1+0)*x^2}=3$$

Can you please tell me where am I doing wrong? Or why these kind of operation doesn't work here? Thank you

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Take a closer look at the zeros in (1-0-0) and (1+0). Note that $x$ to going to 2, not infinity. – Byron Schmuland Nov 11 '11 at 16:18
$x\rightarrow 2$ not to $\infty$ , you have to factor both numerator and denominator.. – pedja Nov 11 '11 at 16:19
Andrew, dividing by the leading term is best when evaluating "end behavior" (i.e. limits at infinity) – The Chaz 2.0 Nov 11 '11 at 16:24
Like Byron said: dividing with the leading term is the strategy for the limit when $x\to\infty$, and doesn't work as well when $x\to 2$. – Willie Wong Nov 11 '11 at 16:25
@Andrew: no, $\frac{x}{3x^2}$ doesn't go to zero as $x \to 2$. It goes to $\frac{2}{12}$ – Ross Millikan Nov 11 '11 at 16:26

For this problem, you'll want to factor the numerator and the denominator,
(which both tend to $0$ as $x \to 2$), cancel a factor of $x-2$, and then try direct substitution again.

$\frac{3x^2 -x - 10}{x^2-4} = \frac{(3x +5)(x - 2)}{(x + 2)(x - 2)} = \frac{3x + 5}{x + 2}$ (when $x \neq 2$)

Now letting $x \to 2$, we get $\frac{3 \cdot 2 + 5}{2 + 4} = \frac{11}{4}$, as desired.

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What if we can't factorize? Should we assume then that the limit DNE? or is 0? – Andrew Nov 11 '11 at 16:33
There are, in fact, at least 8 techniques (not counting L'Hospital's rule) for evaluating 0/0 limits. I say 8, since my AP Calc teacher gave a handout listing them. Sure wish I had that handout! Some of the techniques are: multiplying by radical conjugate, factoring, trig results (sinx/x $\to$ 1 as $x \to 0$, compound fraction/lcd manipulation, ... – The Chaz 2.0 Nov 11 '11 at 16:36
I guess direct substitution should be on the list, since that's a good place to start! – The Chaz 2.0 Nov 11 '11 at 16:40

The original poster asked in a comment: "What if we can't factorize? Should we assume then that the limit DNE? or is 0?"

You had $$\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$$ and as $x\to2$, the numerator and denominator both approach $0$. If the numerator and denominator are polynomials, then:

• If the numerator approaches a non-zero number and the denominator approaches $0$, then the limit does not exist (in some cases it's $\infty$ or $-\infty$, in which case the result is often phrased as "the limit does not exist").
• If the denominator is a non-zero number then just plug in the number that $x$ is approaching (in this case $2$) and that's the limit.
• (The really important case) If they both approach $0$, then use the fact from algebra described below.

Algebra tells us that if you plug a number into a polynomial and get $0$, that tells you something about how to factor it. If you plug $2$ into $3x^2-x-10$ and get $0$, that means $x-2$ is one of the factors. Similarly, you plug $2$ into $x^2-4$ and get $0$, and that tells you $x-2$ is one of the factors. So $$\frac{3x^2-x-10}{x^2-4} = \frac{(x-2)(\cdots\cdots)}{(x-2)(\cdots\cdots)}.$$ Then you need to figure out what goes in place of $(\cdots\cdots)$ in each case. You can do that by long division, dividing $3x^2-x-10$ by $x-2$, and similarly dividing $x^2-4$ by $x-2$. That will work even in cases that would be difficult to factor if you didn't have this way of knowing that $x-2$ is one of the factors. And you don't need to factor completely; you only need to pull out any factors that are $0$ when $x=\text{(in this case) }2$.

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+1 Assuming you were taught polynomial division, which I've found is rarer and rarer in high schools... – user7530 Nov 11 '11 at 20:01