Correct option is (a) 2
The best I can explain: \(\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{e^x+xcos(x)}{sin(x)}=\lim_{x\rightarrow 0}\frac{e^x}{sin(x)}+\lim_{x\rightarrow 0}\frac{xcos(x)}{sin(x)}\)=1/0+0/0, i.e both are indeterminate
i.e., applying L’Hospital Rule
\(\lim_{x\rightarrow 0}\frac{e^x}{sin(x)}=\lim_{x\rightarrow 0}\frac{e^x}{cos(x)}+\lim_{x\rightarrow 0}\frac{cos(x)-xsin(x)}{cos(x)}\)=1+1=2