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