Correct Answer - Option 3 :
\(\frac{1}{2}\)
Concept:
For solving Limit: Remove the indeterminate form from the numerator and denominator.
sin 2x = 2sin x cos x
Calculation:
\(\rm \mathop{\lim}\limits_{x\rightarrow 0}\frac{\sin x}{\sin 2x}\)
\(=\rm \mathop{\lim}\limits_{x\rightarrow 0}\frac{\sin x}{2\sin x\cos x}\)
\(=\rm \mathop{\lim}\limits_{x\rightarrow 0}\frac{1}{2cosx}\)
Putting the limit value x = 0
\(\therefore \rm \mathop{\lim}\limits_{x\rightarrow 0}\frac{sinx}{sin2x}=\frac{1}{2}\)
Hence , option 3 is correct.
