$\lim_\limits{z\to\infty} \sqrt{z-2i} - \sqrt{z-i} ,$ where z is complex no.
How to evaluate this?
I tried by assuming $z = x+iy$ and evaluated $z-2i = x+ i(y-2)$ and $z-1 = x + i(y-1)$ and after putting the value in the given question , I couldn't think of the next step at all
Typically, to find the limit of $\sqrt{a}- \sqrt{b}$ where both a and b go to infinity, one multiplies by the "unit fraction" $\frac{\sqrt{a}+ \sqrt{b}}{\sqrt{a}+ \sqrt{b}}$. Since $(\sqrt{a}- \sqrt{b})(\sqrt{a}+ \sqrt{b})= a- b$ the fraction becomes $\frac{a- b}{\sqrt{a}+ \sqrt{b}}$.
In this problem, $a= z- 2i$ and $b= z+ i$ so $a- b= z- 2i- z- i= -3i$. The fraction is now $\frac{3i}{\sqrt{z- 2i}+ \sqrt{z+ i}}$. As z goes to infinity, the denominator goes to infinity and, since the numerator is constant, the fraction goes to 0.