Question

Need help in b and c. show calculation pls.​

Answer 1

√(16 - x^2) is defined only for -4 ≤ x ≤ 4, and is continuous over this domain, so

$\displaystyle\lim_{x\to-4^+}\sqrt{16-x^2}=\sqrt{16-(-4)^2}=0$

From the other side, the limit does not exist, because all x < -4 do not belong to the domain.

Taken together, the two-sided limit also does not exist.