Proving convergence Using the definition Use the definition of convergence to prove if $x_n$ converges to $5$, then 
$\frac{x_n+1}{\sqrt{x_n-1}}$ converges to $3$.
 A: There is an almost automatic but not entirely pleasant procedure. Let $y_n=\frac{x_n+1}{(x_n-1)^{1/2}}$.  We will need to examine $|y_n-3|$.
Note that
$$y_n-3=\frac{x_n+1}{(x_n-1)^{1/2}}-3=\frac{(x_n+1)-3(x_n-1)^{1/2}}{(x_n-1)^{1/2}}.$$
Multiply top and bottom by $(x_n+1)+3(x_n-1)^{1/2}$. This is the usual "rationalizing the numerator" trick.
After some manipulation we get
$$y_n-3=\frac{(x_n-2)(x_n-5)}{(x_n-1)^{1/2}((x_n+1)+3(x_n-1)^{1/2}    )}.\tag{1}$$
Now we need to make some estimates. Ultimately, given $\epsilon\gt 0$, we will need to choose $N$ so that if $n\ge N$ then $|y_n-3|\lt \epsilon$.
Observe first that there is an $N_1$ such that if $n\ge N_1$ then $|x_n-5|\lt 1$. That means that if $n\ge N_1$, we have $4\lt x_n\lt 6$.
In particular, $(x_n-1)^{1/2}\gt \sqrt{3}\gt 1$, and  $(x_n+1)+3(x_n-1)^{1/2}\gt 8$. Also, $|x_n-2|\lt 4$. It follows from Equation (1) that if $n\ge N_1$, then
$$|y_n-3| \lt \frac{4}{(1)(8)}|x_n-5|.$$
By the fact that the sequence $(x_n)$ converges, there is an $N_2$ such that if $n\ge N_2$ then $|x_n-5|\lt \frac{(1)(8)}{4}\epsilon$.
Finally, let $N=\max(N_1,N_2)$. If $n\ge N$, then, putting things together, we find that 
$$|y_n-3|\lt \epsilon.$$
Remark: We were using only the definition of convergence. If we relax and use some theorems, all we need to say is that the function $\frac{x+1}{(x-1)^{1/2}}$ is continuous at $x=5$. 
