c. \mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt{cosx}}{tan^2x}=\mathop {\lim }\limits_{x \to 0}\frac{1-cosx}{tan^2x(1+\sqrt{cosx})}=$\mathop {\lim }\limits_{x \to 0}\frac{sin^2\frac{x}{2}.cos^2x}{sin^2x(1+\sqrt{cosx})}=\mathop {\lim }\limits_{x \to 0}\frac{(\frac{1}{4})x^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2cos^2x}{sin^2x(1+\sqrt{cosx})}=\frac{1}{8}$
c. \mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt{cosx}}{tan^2x}=$\mathop {\lim }\limits_{x \to 0}\frac{1-cos^2x}{tan^2x(1+\sqrt{cosx})}=\mathop {\lim }\limits_{x \to 0}\frac{sin^2x.cos^2x}{sin^2x(1+\sqrt{cosx})}=\mathop {\lim }\limits_{x \to 0}\frac{cos^2x}{1+\sqrt{cosx}}=\frac{1}{2}$
c.
\mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt{cosx}}{tan^2x}=
\mathop {\lim }\limits_{x \to 0}\frac{1-cosx}{tan^2x(1+\sqrt{cosx})}=$\mathop {\lim }\limits_{x \to 0}\frac{sin^2
\frac{x
}{2}.cos^2x}{sin^2x(1+\sqrt{cosx})}
=\mathop {\lim }\limits_{x \to 0}\frac{
(\frac{1}{4})x^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2cos^2x}{
sin^2x(1+\sqrt{cosx}
)}=\frac{1}{
8}$