6
$\begingroup$

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?

  • 1
    Questions should be self-contained...2011-08-17
  • 2
    Yes, the question should be rewritten. But the function is not continuous at $0$, because its value at $0$ is $0$, and the function tends to $\frac{\pi}{2}$ as $t$ tends to 0 from above. For a function $f$ to be continuous at $a,$ we need $\lim_{t \to a} f(t) = f(a).$2011-08-17

3 Answers 3