$\lim_{n\to\infty}\frac{e^{\frac{1}{n^2}}-\cos \frac{1}{n}}{\frac{1}{n}\log(\frac{n+1}{n})-(\sin\frac{2}{n})^2}$ $$\lim_{n\to\infty}\frac{e^{1/n^2}-\cos \frac{1}{n}}{\frac{1}{n}\log(\frac{n+1}{n})-(\sin\frac{2}{n})^2}, \quad n \in \mathbb{N}$$


I'm stuck trying to evaluate this limit. I have tried applying l'Hôpital's rule to the corresponding function in $\mathbb{R}$, but even after differentiating all that stuff I still end up with a $\frac{0}{0}$ form. There has to be something I'm missing. Any hints?
Hint. Try by using Taylor expansions. As regards the numerator
$$e^{\frac{1}{n^2}}-\cos \frac{1}{n}=1+\frac{1}{n^2}+o(1/n^2)-
\left(1-\frac{1}{2n^2}+o(1/n^2)\right)=\frac{3}{2n^2}+o(1/n^2).$$
Whereas, for the denominator we have that
$$\frac{1}{n}\log(1+\frac{1}{n})-(\sin\frac{2}{n})^2=?$$
Can you take it from here?
Hint

Set $1/n=h$. to find $$\dfrac{(e^{h^2}-1)/h^2+(1-\cos h)/h^2}{\ln(1+h)/h-4(\sin2h/2h)^2}$$

Use $\lim_{x\to0}\dfrac{\sin x}x=1$

$\lim_{x\to0\dfrac{e^x-1}x=1=\lim_{x\to0}\dfrac{\ln(1+x}x$