Răspuns:
Explicație pas cu pas:
[tex]-2\leq 0,~deci,~f(x)=4^x.~Atunci~ \lim_{x \to -2} f(x)=f(-2)=4^{-2}=\dfrac{1}{4^2}=\dfrac{1}{16}.\\E~necesar~de~calculat~f(0-0),~f(0),~f(0+0):f(0-0)= \lim_{x \to 0-0,~x<0} f(x)=4^0=1.\\f(0)=\lim_{x \to 0} f(x)=4^0=1.\\f(0+0)=\lim_{x \to 0+0,~x>0} f(x)=log_4(0+0)=- \infty.\\Deoarece~f(0-0)=f(0)\neq f(0+0),~=>~functia~nu~are~limita~in~0.\\4>0,~deci~f(x)=log_4x,~=>~ \lim_{n \to 4} f(x)=log_44=1.[/tex]
