Find the constant $c$ if you have a given limit 
If $\displaystyle\lim_{x\to11}\frac{|x^2-6x|-|x^2+cx|}{x-11}.=22$, find the value of the constant $c$.

I have a problem getting the correct result which is $-16$. I tried absolutely everything and the closest I got was $-6$. My idea was to just check limits from both sides (although one side would be enough because I was sure an answer is a final number, thus I chose $11$ from the positive side so the absolute values cancel out easier). When I simplify as much as I can, I really don't get anywhere. I don't know what to do with the $x-11$ in the denominator.
 A: Note that near $x=11$ one has $x^2-6x>0$ so $|x^2-6x|=x^2-6x$. If we also had $x^2+cx>0$ near $x=11$, then the numerator would simplify to $-(6+c)x$ and the limit of the entire fraction would not exist. Therefore $c$ must be such that $x^2+cx<0$ near $x=11$. Then the original expression is equal to
$$
\frac{2x^2-x(6-c)}{x-11}=\frac{x(2x-(6-c))}{x-11}.
$$
The singularity will cancel out only if $2x-(6-c)=0$ for $x=11$.
Thus
$$
22-(6-c)=0
$$
and solving for $c$ we get $c=-16$.
A: First, as $x$ is close to $11$ you have that $x^2-6x>0$, so your expression is
$$
\frac{x^2-6x-|x^2+cx|}{x-11}.
$$
As $x-11\to0$, the numerator needs to be or approach a multiple of $x-11$. In particular, it goes to zero as $x\to11$. So
$$
0=\lim_{x\to 11} x^2-6x-|x^2+cx|=121-66-|121+11c|=55-|121+11c|
$$
This requires
$$
|121+11c|=55. 
$$
So either
$
-121-11c=55
$, giving $c=-16$, or
$121+11c=55$, giving $-6$. The second choice makes the numerator identically zero, so the limit would be zero. That leaves $c=-16$.
A: As an exploratory tactic,

*

*the left hand side would be $O(x)$ if $c \le -11$


*and the right hand side looks like $2 \times 11$


*so I might start by considering whether there was a $c$ which gave $x^2-6x +x^2+cx = 2x(x-11)$


*and there is one, namely $c=-16$.  Checking this, it works.
Looking for other solutions

*

*If $c>-11$ then the left hand side is the limit of $\frac{x^2-6x-x^2-cx}{x-11} = \frac{-x(6+c)}{x-11}$


*which cannot have a finite limit as $x \to 11$ when $c\not=-6$


*and has a zero limit when $c=-6$.


*So that fails and there is no other solution
