Correct Answer - Option 3 : 2 cos 1
Concept:
\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x }{x} =1\)
Trigonometry formulas:
\(\rm \sin C - \sin D = 2\sin \frac{C-D}{2}\cos \frac{C+D}{2}\)
\(\rm \sin C + \sin D = 2\sin \frac{C+D}{2}\cos \frac{C-D}{2}\)
Calculation:
Here, we have to find the value of the limit \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin (1 + x) - \sin (1-x)}{x}\)
As we know, \(\rm \sin C - \sin D = 2\sin \frac{C-D}{2}\cos \frac{C+D}{2}\)
So, \(\rm \sin (1 + x) - \sin (1-x) = 2\sin \frac{(1 + x)-(1-x)}{2}\cos \frac{(1 + x)+(1-x)}{2} =2\sin x \cos 1\)
Now, \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin (1 + x) - \sin (1-x)}{x}\)
= \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{2\sin x \cos 1}{x}\)
= 2 cos 1 × \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x }{x} \)
= 2 cos 1 × 1
= 2 cos 1
