A) 1/4
B) 3/4
C) 1/2
D) 1
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(\tan x)}^{3/2}}[1-{{(\cos x)}^{3/2}}]}{{{x}^{3/2}}.{{x}^{2}}}\] \[=1\times \underset{x\to 0}{\mathop{\lim }}\,\frac{1-{{\cos }^{3}}x}{{{x}^{2}}}.\frac{1}{1+{{(\cos x)}^{3/2}}}\] \[=\frac{1}{2}.\frac{1}{2}(1+\cos x+{{\cos }^{2}}x)=\frac{3}{4}\]You need to login to perform this action.
You will be redirected in
3 sec