Overdamped and critically damped 
Consider $y''+2by'+w^2y=0$. Show that as the limit of $b\to w$, the overdamped solution is equal to the critically damped solution.

The roots are $D=-b\pm\sqrt{b^2-w^2}$. Thus, if $b^2>w^2$ then it is overdamped and if $b^2=w^2$ it is critically damped. 
Now, for overdamped we have, $y=Ae^{-\lambda t}+Be^{-\mu t}$, where $\lambda=b+\sqrt{b^2-w^2}$ and $\mu=b-\sqrt{b^2-w^2}$.
For critically damped, we have, $y=(A+Bt)e^{-bt}$. 
Now, if I take $\lim_{b\to w}$ then $\lambda=w, \ \mu=w$, thus $y=(A+B)e^{-wt}$ but that doesn't equal to $y=(A+Bt)e^{-bt}$.
 A: Write it as
$$y=Ae^{-bt}e^{-\sqrt{b^2-w^2}t}+Be^{-bt}e^{+\sqrt{b^2-w^2}t}$$
$$y=e^{-bt}(Ae^{-\sqrt{b^2-w^2}t}+Be^{+\sqrt{b^2-w^2}t})$$
This looks promising. Because the exponent tends toward $0$ in the limit, you can use Taylor expansion:
$$y=e^{-bt}(A-A\sqrt{b^2-w^2}t+\frac{A}{2}(\sqrt{b^2-w^2}t)^2+\cdots+$$
$$+B+B\sqrt{b^2-w^2}t+\frac{B}{2}(\sqrt{b^2-w^2}t)^2+\cdots)$$
$$y=e^{-bt}((A+B)-(A-B)\sqrt{b^2-w^2}t+\frac12(A+B)(b^2-w^2)o(t^2))$$
The important part to realize is, that you can't treat $A$ and $B$ as constants during the limit. The solution of the DE depends on the initial conditions, so the $A$ and $B$ for fixed boundary conditions depend on $b$ and $w$. You can do this explicitly by yourself. Just invent some initial condition $u=y(0)$ and $v=y'(0)$ and solve it, then take the limit.
However, simple counting of degrees of freedom above tells you what happens. You can always choose $A$ and $B$ such that $A-B$ tends to infinity in a way that cancels out the $\sqrt{b^2-w^2}$ in the limit. That's exactly what happens if you fix the initial condition. However, the quadratic term has the same $A+B$ behaviour as the constant term... so, this one has to go to $0$ (otherwise $A+B$ would need to diverge, and that makes the entire solution divergent).
You can also write the initial solution as a combination of hyperbolic sine and cosine. There it is more obvious that $\cosh(x)\to 1$ and $\sinh(x)\to x$ for small arguments. This also works from the other side, if $b$ goes to $w$ from below (you get $\cos(x)\to 1$ and $\sin(x)\to x$ then).

Edit:
The limiting process would go as such. Let $y(0)=u$ and $y'(0)=v$. These mean
$$y(0)=u=A+B$$
$$y'(0)=v=-\lambda A-\mu B=-\frac12(\lambda+\mu)(A+B)-\frac12(\lambda-\mu)(A-B)$$
This seems like black magic, but it's just a shorter way instead of expressing $A$ and $B$ with $u$ and $v$.
So, express:
$$A+B=u$$
$$A-B=\frac{-v-\frac12u(\lambda+\mu)}{\frac12(\lambda-\mu)}=\frac{-v-bu}{\sqrt{b^2-w^2}}$$
The Taylor expanded solution becomes:
$$y=e^{-bt}((A+B)-(A-B)\sqrt{b^2-w^2}t+\frac12(A+B)(b^2-w^2)o(t^2))=$$
$$=e^{-bt}(u-\frac{-v-bu}{\sqrt{b^2-w^2}}\sqrt{b^2-w^2}t+\frac12u(b^2-w^2)o(t^2))$$
$$=e^{-bt}(u+(v+bu)t+\frac12u(b^2-w^2)o(t^2))$$
You see, now if you take the limit $b\to w$, the linear term survives.
