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Alright so I had a math question one on of my tests yesterday and it followed.

Choose a value of a so that this limit has a definition (since if $x \to 2$, $x^2-4 = 0$)

$$\lim_{x \to 2} \frac{x^2 + ax + 6}{x^2-4}$$

I thought I was supposed to use the variable $a$ to go back with the conjugate rule but I'm not sure anymore. I got it to $-5$.

Since the question is pretty specific I couldn't find the answer anywhere. (Sorry if the tag wasn't the best, first question so had no rep for limit)

Thanks in advance!

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    $\begingroup$ what can you factorize the denominator to? If you do that, is part of the factorisation problematic at x=2? - how could you get rid of that problem? (by cancellation possibly) $\endgroup$ – Cato Oct 14 '16 at 13:43
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Since $x^2-4\rightarrow 0$ as $x\rightarrow 2$, it is necessary that $$x^2+ax+6\rightarrow 0\text{ as }x\rightarrow 2$$ as well, otherwise the limit would not be finite. Since $P(x)=x^2+ax+6$ is continuous, this amounts to $P(2)=0$.

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  • $\begingroup$ once you've found the 'a' that makes the numerator equal to zero at x=2, you've not finished until you've shown that the limit exists at x=2 $\endgroup$ – Cato Oct 14 '16 at 13:48
  • $\begingroup$ I know that, which is why I wrote "it is necessary that". $\endgroup$ – Olivier Moschetta Oct 14 '16 at 13:52
  • $\begingroup$ Thank you, couldn't wait for the test to get back :) $\endgroup$ – Lee Dash Oct 14 '16 at 13:58
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As the denominator that is $$ x^2 - 4 \rightarrow 0 $$ . Hence for limit to exist the numerator ie $$x^2 + ax + 6 \rightarrow 0 $$ because then only we can apply the L'Hopitals method to find the limit . Hence $$ x^2 + ax+ 6 = 0$$ at $$ x= 6$$ Solve for a and you'll get $a= -5$

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    $\begingroup$ Ahh alright thanks boys! $\endgroup$ – Lee Dash Oct 14 '16 at 13:57
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since the denominator factorises

$x^2 - 4 = (x - 2)(x+2)$

and $x-2$ is going to make the denominator 0 at $x = 2$, we want to seek to cancel the $x-2$

so we want

$x^2 + ax + 6$ to have $(x - 2)$ as a factor, which can be achieved by setting

$x^2 + ax + 6 = 0$ at x = 2 and solving

to give a = -5

after which

$(x^2 - 5x + 6) / (x^2 - 4) $

$= (x - 2)(x-3) / [(x + 2)(x-2)] $

$= (x - 3)/(x +2) $

where the limit exists at $x = 2$, and is $-1/4$

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