Within a limit, find the constant c.

Question

Find the constant $c$ such that the limit of the following, exists.

$$\lim_{x\to 3} \frac{x^2+x+c}{x^2-5x+6}$$

What I've tried

So typically to find a limit, you substitute the number that x approaches, to find the limit, but given the above equation, that can't be done immediately as there are two asymptotes at 2 and 3 -as I know that the factoring of $(x^2 + -5x + 6)$ gives us $(x-2)(x-3)$, but from there, I'm fairly lost. Some help would be greatly appreciated.

Answer 1

In order for the limit to exist you need to get rid of the asymptote that occurs when $x=3$, caused by the denominator being $0$. As you know, you can factorise the denominator as

$$(x-2)(x-3)$$

Hence you only need to find the $c$ so that the $(x-3)$ term can be cancelled, since then the denominator will no longer become $0$ at $x=3$. Hence, find the constant $c$ so that

$x^2 + x + c$ has $x-3$ as a factor

Answer 2

Since $\lim_{x\to3}x^2-5x+6=0$ and since the limit$$\lim_{x\to3}x^2+x+c\tag1$$always exists (it is equal to $c+12$), if the limit$$\lim_{x\to3}\frac{x^2+x+c}{x^2-5x+6}\tag2$$exists, then $(1)=0$; in other words, $c=-12$. Now, prove that, in that case, the limit $(2)$ exists indeed.

Answer 3

Hint

to remove the $(x-3)$ factor you can say, that there exists $a$, such that: $$x^2+x+c=(x-3)(x-a)$$

Answer 4

Use algebra of limits directly. We have $$c=\lim_{x\to 3}(x^2+x+c)-(x^2+x)=\lim_{x\to 3}(x^2-5x+6)\cdot\frac{x^2+x+c}{x^2-5x+6}-12=0\cdot l-12=-12$$