The improper integral $\int_{0}^{x}y^{-1/2}dy$ is
$1.$ Continuous on $[0,\infty).$
$2.$ Continuous on $(0,\infty).$
$3.$ Continuous only on $[1/2,\infty)$
It is cleat that option $3$rd is wrong one. I am confused about $1$st and $2$nd option. I tried it as $$\int_{0}^{x}y^{-1/2}dy=2\sqrt{x}$$ so it is continuous on $[0,\infty).$ But i am confused about the point that the function $\int_{0}^{x}y^{-1/2}dy$ is not defined at $0$ so not continuous at $0.$ Please suggest me . Thanks a lot.
Your confusion was leading you to the correct answer: $2\sqrt{x}$ is not defined at $x<0$ so the limit from the left does not exist so that integral cannot be continuous at $x=0$. Then number 2 is the right answer.
This is the answer in real analysis, where the domain to be considered is the netire real line and you have to care about the limit from the left of zero. If your universe were $\Bbb{R}^+ \cup 0$ then you would have to say that the function is continuous everywhere.