Continuity of function with parameter I have this function. I need to find all the parameters a (real number) so that the function is continuous.
$${f(x)=\begin{cases}
sin(ax-a)/(x-1) & \text{, } x< 1 \\
(a/(x+2))+ax & \text{, } x\geq1
\end{cases}} \\
$$
When I substitute first I get a limit of type 0/0. If I extend the expression by a/a and make adjustments I get as a result a. For the second limit I get a/3 + a. So I get a = a / 3 + a which is probably not good. How should I work with the first limit?
 A: The language you use to describe what you're doing isn't the most accurate, but it is true that
$$\lim_{x \rightarrow 1^-} f(x) = a,$$
and
$$\lim_{x \rightarrow 1^+} f(x) = (a/3) +a,$$
and continuity at $1$ requires that
$\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^+} f(x) = f(1)$.
So the restriction you found on $a$ is correct. And it is a possibility that there are no values of $a$ that will make $f$ continuous, or maybe there is a value but it would make $f$ very boring.
(This is more a point about what kind of questions teachers and text book writers tend to create for us, than it is a point about what kind of math problems exist in the real world. Our classes seem to give us problems where finding solutions involves integers or small denominator fractions, and answers are of middle complexity, not super complicated and not boringly simple. But hopefully we build up our skills so that when the answer is boring we are confident enough to say "Yep, the answer here looks suspicious but it's right.".)
A: So if we have a limit from right and left, I finally get the equation
a + a / 3 = a.
and the result is a = 0 ?
