limit as x→3 from the left of Square root (9-x^2)
Notice that $\displaystyle y = \sqrt{9 - x^2}$
$\displaystyle y^2 = 9 - x^2$
$\displaystyle x^2 + y^2 = 3^2$.
This is a circle of radius $\displaystyle 3$ centred at $\displaystyle (0,0)$.
Therefore its domain is $\displaystyle x \in [-3, 3]$.
Since $\displaystyle x$ is defined approaching $\displaystyle 3$ from the left, we can simply substitute $\displaystyle x = 3$ into the equation.
So $\displaystyle \lim_{x \to 3}\sqrt{9 - x^2} = \sqrt{9 - 3^2} = \sqrt{0} = 0$.