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!
3 Answers
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 $P(2)=0$.
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 $a= -5$
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 $-1/4$
