Объяснение:

[tex]a)\ \lim\limits_{x \to 0} \frac{5\sqrt{x} -x}{x-4\sqrt{x} } =\lim\limits_{x \to 0} \frac{\sqrt{x} *(5-\sqrt{x} )}{\sqrt{x} *(\sqrt{x} -4) } =\lim\limits_{x \to 0} \frac{5-\sqrt{x} }{\sqrt{x} -4 } =\frac{5-\sqrt{0} }{\sqrt{0}-4 }=\frac{5-0}{0-4} =\frac{5}{-4} =-\frac{5}{4}.\\ b)\ \lim\limits_{x \to \frac{\pi }{12} }sin4x=sin\frac{4x*\pi }{12} =sin\frac{\pi }{3} =\frac{\sqrt{3} }{2} .\\[/tex]

[tex]c)\ \lim\limits_{x \to -4}\frac{x^2+2x-8}{x^2-16}= \lim\limits_{x \to -4}\frac{x^2+4x-2x-8}{x^2-4^2}=\lim\limits_{x \to -4}\frac{x*(x+4)-2*(x+4)}{(x-4)*(x+4)} =\\=\lim\limits_{x \to -4}\frac{(x+4)*(x-2)}{(x-4)*(x+4)} =\lim\limits_{x \to -4}\frac{x-2}{x-4}=\frac{-4-2}{-4-4}=\frac{-6}{-8} =\frac{6}{8}=\frac{3}{4}.[/tex]