Correct Answer - Option 2 :
\(\frac{1}{2}\)
Calculation:
We have,
\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\tan x\, -\, \sin x}{\sin^{3}x}\)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\frac{\sin x}{\cos x}\, -\, \sin x}{\sin^{3}x}\) (∵ tan x = sin x/cos x)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\sin x(1-\cos x)}{\cos x\, \sin^{3}x}\)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{(1-\cos x)}{\cos x\, \sin^{2}x}\)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{(1-\cos x)}{\cos x\, (1-\cos^{2}x)}\) (∵ sin2 x = 1 - cos2 x)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{(1-\cos x)}{\cos x\, (1+\cos x)(1-\cos x)}\)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{1}{\cos x\, (1+\cos x)}\)
\(\therefore \frac{1}{2}\) is the correct limit.
