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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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    I would like to be the 9th person to say $x \to 2$, not $x \to \infty$.2011-11-11

2 Answers 2

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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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    I guess *direct substitution* should be on the list, since that's a good place to start!2011-11-11
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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...2011-11-11