Limit with Floor Function

Carefully observe that we have to find out the left hand limit of the function $\large f(x)=\frac{\lfloor{\sqrt{x}} \rfloor ^2}{x}$ . Now, since $x\to 25^{-}$ , so think of a value slightly less than $25$ . Its square root is then slightly less than $5$ . Then, by definition of the floor function, we have $\lfloor{\sqrt{x} \rfloor}=4$ . Then, just proceed as usual by substituting $x=25$ in the denominator of the limit $\displaystyle \large \lim_{x\to 25^{-}} f(x)$ .

$\displaystyle \large \lim_{x\to 25^{-}} f(x)$

$= \displaystyle \large \lim_{x\to 25^{-}} \frac{\lfloor{\sqrt{x}} \rfloor ^2}{x}$

$= \frac{{4}^2}{25} = \frac{16}{25} = \boxed{0.64}$