Question

Find its:-
$\underset{x\to 0}{lim}\frac{{\left(\cos x\right)}^{1/2}-{\left(\cos x\right)}^{1/3}}{{\sin }^{2}x}$

Answer 1

Solution -

${lim}_{x\to 0}\frac{(cosx{)}^{1/2}-(cosx{)}^{1/3}}{si{n}^{2}x}$

Apply L - Hospital rule

${lim}_{x\to 0}\frac{-1/2(cosx{)}^{-1/2}sinx+\frac{1}{3}(cosx{)}^{-2/3}sinx}{2sinxcosx}$

$=\frac{\frac{-1}{2}+\frac{1}{3}}{2}$

$=\frac{-1}{12}$