Correct option is (a) 1
Best explanation: \(\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{tan(x)}{e^x-1}=0/0\) (Indeterminate Form)
By L’Hospital’s rule
\(\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)}=\lim_{x\rightarrow 0}\frac{sec^2(x)}{e^x}\)=1