Calculate the limit: $\lim_{n\rightarrow 0}\frac{\sin(2n)}{\ln(1+n)}$ Calculate the limit with L'Hôpital:
  
  $$\lim_{n\rightarrow 0}\frac{\sin(2n)}{\ln(1+n)}$$


I get $2$ as limit but not sure because I tested with calculator and I get some small thing like $0.032\ldots$

$$(\sin(2n))'=2\cos(2n)$$

$$(\ln(1+n))'=\frac{1}{1+n}$$

$$\lim_{n\rightarrow 0}\frac{2\cos(2n)}{\frac{1}{1+n}}= \lim_{n\rightarrow 0}2\cos(2n) \cdot(1+n)=\lim_{n\rightarrow 0}2\cos(2n)+2n\cdot \cos(2n)=$$

$$2\cdot \cos(0)+0\cdot \cos(0)=2\cdot1+0\cdot1=2$$
Yeah, you're doing right. When you check the result, please make sure that you calculate the angle using radian mode.
$$\lim_{n\rightarrow 0}{\sin(2n)\over \ln(1+n)}$$
$$=\lim_{n\rightarrow 0}{\sin(2n)\over \ln(1+n)}{2n\over 2n}$$
$$=\lim_{n\rightarrow 0}{2n\over \ln(1+n)}$$
$$=\lim_{n\rightarrow 0}\large{2\over {\ln(1+n)\over n}} = 2$$