About this limit, $\lim_{t\to0^+} \sum_{n=1}^\infty \frac{\sqrt t}{1 + tn^2}$ This recent question, Evaluating a limit, $\lim\limits_{t\rightarrow 0^{+}} {\sum\limits_{n=1} ^{\infty} \frac{\sqrt{t}}{1+tn^2}}$, asked for the value of
$$\lim_{t\to0^+} \sum_{n=1}^\infty \frac{\sqrt t}{1 + tn^2}$$
So that I could better understand the answer can someone explain if this function of $t$ is discontinuous at $t=0$ and that is why the right-sided limit has to be taken? Does this function have any significance?  
 A: The function $\sqrt{t}$ is not defined for negative $t$. So it makes no sense to look at the behaviour as $t$ approaches $0$ through negative values.
The sum, as a function $F(t)$, can be expressed, for $t>0$, in terms of the hyperbolic cotangent of $\pi/\sqrt{t}$. For $t=0$, each term is $0$, so $F(0)=0$. 
It was shown by @Shai Covo and @anon that $\lim_{t\to 0+}F(t)=\pi/2$.
So $F(t)$ is not continuous (from the right) at $t=0$. 
However, if we define the function $G(t)$ by $G(t)=F(t)$ if $t>0$, and $G(0)=\pi/2$, then the function $G$ is continuous (from the right) at $t=0$.
This is because $\lim_{t\to 0+}F(t)=\pi/2$.  
Of course $G(t)$ cannot be fully continuous at $0$, since we have not even defined it for negative $t$.  If we cared to, we could define $H(t)$ by $H(t)=G(t)$ if $t\ge 0$, and $H(t)=G(|t|)$ for $t<0$.  Then $H(t)$ would be continuous for all $t$.  This is not all that unreasonable.  Instead of summing $\sqrt{t}/(1+n^2 t)$, we would be summing $\sqrt{|t|}/(1+n^2 |t|)$.
The discontinuity (from the right)  of $F(t)$ at $t=0$ is in a sense not a mathematically significant one.  The technical term is that it is a removable discontinuity. The value of $F(0)$ is the "wrong one" for continuity from the right, but that can be easily changed by replacing that value by the "correct one," which should be $\pi/2$.
A: The "significance" is that the sum is actually a Riemann sum that approximates the integral $\int_0^\infty \frac{dx}{1+x^2}$ with interval lengths $\sqrt{t}$.  Consequently as $t\to 0+$, the sum approaches the integral.
It is in fact discontinuous from the right.  If $t$ is actually equal to $0$, then the sum is exactly $0$, but as $t$ approaches $0$, the sum approaches $\pi/2$, not $0$.  That is a discontinuity.  But that does not explain why the limit is one-sided.  What one would do with $\sqrt{t}$ if $t$ were negative is not altogether clear, and at any rate a negative number cannot be the length of the intervals in a Riemann sum.
A: To make the notation simpler, substitute $t\mapsto t^2$:
$$
\lim_{t\to0^+}\sum_{n=1}^\infty\frac{\sqrt t}{1+tn^2}
=\lim_{t\to0^+}\sum_{n=1}^\infty\frac{t}{1+{(tn)^2}}
$$
First note that
$$
\begin{align}
\sum_{n=N/t}^\infty\frac{t}{1+{(tn)^2}}
&\le\frac1t\sum_{n=N/t+1}^\infty\frac1{n^2}\\
&\le\frac1t\sum_{n=N/t+1}^\infty\frac1{n(n-1)}\\
&=\frac1N
\end{align}
$$
Since Riemann integrals are only over finite intervals, we need to restrict the sum to get the Riemann Sum (where $tn=x$ and $t=\mathrm{d}x$)
$$
\begin{align}
\lim_{t\to0^+}\sum_{n=1}^{N/t}\frac{t}{1+{(tn)^2}}
&=\int_0^N\frac{\mathrm{d}x}{1+x^2}\\
&=\arctan(N)
\end{align}
$$
Therefore, letting $N\to\infty$,
$$
\begin{align}
\lim_{t\to0^+}\sum_{n=1}^\infty\frac{t}{1+{(tn)^2}}
&=\arctan(N)+O\left(\frac1N\right)\\
&=\frac\pi2
\end{align}
$$

Alternate approach
Set $t=1/k$, then
$$
\lim_{t\to0^+}\sum_{n=1}^\infty\frac{\sqrt t}{1+tn^2}
=\lim_{t\to0^+}\sum_{n=1}^\infty\frac{t}{1+{(tn)^2}}
=\lim_{k\to\infty}\sum_{n=1}^\infty\frac{1/k}{1+{(n/k)^2}}
$$
Since $\frac{1/k}{1+{(n/k)^2}}$ is monotonically decreasing (in $n$), the Integral Test says
$$
\int_0^\infty\frac{\mathrm{d}x}{1+x^2}
\le\sum_{n=1}^\infty\frac{1/k}{1+{(n/k)^2}}
\le\int_{1/k}^\infty\frac{\mathrm{d}x}{1+x^2}
$$
By the Squeeze Theorem, we get
$$
\lim_{k\to\infty}\sum_{n=1}^\infty\frac{1/k}{1+{(n/k)^2}}
=\int_0^\infty\frac{\mathrm{d}x}{1+x^2}
=\frac\pi2
$$
