Continuous dependence of zeros on a parameter Let $F:I\times J\to\mathbb{R}$ be a $C^k$ (or analytic) function, with $I,J$ real open intervals.
Set $f_\lambda(x):=F(\lambda,x)$ and consider the parametric equation
$$f_\lambda(x)=0\,.$$
Assume its solutions are at least $1$ and at most $n$ (not dependent on $\lambda$) and denote them by $$x_1(\lambda)\leq x_2(\lambda)\leq\dots\leq x_n(\lambda)\,.$$
Is it possible to say that $\lambda\mapsto x_1(\lambda)$ is a continuous function?
Further conditions on $F$ are required?
Note. For the sake of semplicty one may start with $n=2$. In this case there could be for example a value $\lambda_0$ such that for $\lambda\leq\lambda_0$ the equation $f_\lambda(x)=0$ has 1 solution $x_1(\lambda)$, while for $\lambda>\lambda_0$ it has $2$ different solutions $x_1(\lambda)<x_2(\lambda)$.
 A: The answer to your question is no. It is not possible to claim that $\lambda \mapsto x_1(\lambda)$ is a continuous function. To show it, it is enough to provide a counterexample.
Let us consider $I=J=(0,1)$ for simplicity. Then we define the polynomial function:
$$F(x,\lambda):= -\frac{1}{3}x^3 +\frac{1}{2}x^2 -\frac{2}{9}x +\frac{\lambda}{20}.$$
Observe that, since it is a polynomial, $F$ is analytic, so $\mathcal{C}^\infty$ and so on. I apologize for the "weird" numbers, but I wanted to show exactly what is going on with pictures, and Mathematica essentially forced them on me. 
This is the picture of $F(x,0.4)$:
 
This is the picture of $F(x,0.6)$:

And finally this is the picture of $F(x,0.8)$:

You see that, while $\lambda$ is getting bigger, we are shifting the graph of $F$. And it is also clear that, for a specific $\lambda$ in the interval $(0.6,0.8)$, your function $x_1(\lambda)$ will jump to the right. It will assume those values that you were calling $x_3(\lambda)$ until an $\epsilon$ before. 
This answer is "sketchy" on purpose, I think that it gives you exactly the idea of the kind of things that can go wrong. 
