Limit of the sequence using intermediate value theorem $\lim_{x\to\infty}[\frac{1}{\sin(\frac{1}{x+\psi(x)})}-\frac{1}{\sin(\frac{1}{x+\phi(x)})}]$ Find the limit of:
$$\lim_{x\to\infty}[\frac{1}{\sin(\frac{1}{x+\psi(x)})}-\frac{1}{\sin(\frac{1}{x+\phi(x)})}]$$
where:
$$\psi(x)=(1+\frac{1}{x})^x, ~~~~~~~\phi(x)=\sqrt[x]{x}$$
I used Lagrange theorem for the intermediate value for $f(x)=\frac{1}{\sin(x)}$, which is a recommended way of solving the problem, but I am stuck now and I would ask you for some help.
$$\frac{f(b)-f(a)}{b-a}=f'(c),~~~~c\in(a,b)\\ f(b)-f(a)=f'(c)\cdot(b-a) \\ \lim_{x\to\infty}\frac{\frac{1}{\sin(\frac{1}{x+\psi(x)})}-\frac{1}{\sin(\frac{1}{x+\phi(x)})}}{x+\psi(x)-x-\phi(x)}=\lim_{x\to\infty}\frac{\cos(\frac{1}{c})}{\sin^2(\frac{1}{c})\cdot x^2}\cdot (\psi(x)-\phi(x))$$
That's the end of my story. I have also tried $\frac{1-\cos(2\alpha)}{2}=\sin^2(\alpha)$, but it leaded me to nowhere. I would really appreciate your hints.

Assuming that $c\approx x$, I have used Taylor expansion for the:
$$ \frac{\cos(\frac{1}{c})}{\sin^2(\frac{1}{c}) \cdot x^2} =\\$$
$$\frac{1-\frac{1}{2x^2}+\frac{1}{4! x^4} + o(\frac{1}{x^4})}{(\frac{1}{x^2}-\frac{1}{3 x^4} + \frac{1}{36 x^6}+o(\frac{1}{x^6}))\cdot x^2} =\\$$
$$\frac{1-\frac{1}{2x^2}+\frac{1}{4! x^4}  +o(\frac{1}{x^4})}{1-\frac{1}{3 x^2} + \frac{1}{36 x^4}+o(\frac{1}{x^4})}$$
After division by $\frac{1}{x^4}$ and taking out $x^4$, the limit is:
$$ \frac{x^4}{x^4}\cdot\frac{(1-\frac{1}{2x^2}+\frac{1}{4!x^4})}{(1-\frac{1}{3x^2}+\frac{1}{36x^4})} = \frac{1-\frac{1}{2x^2}+\frac{1}{4!x^4}}{1-\frac{1}{3x^2}+\frac{1}{36x^4}}$$
for $x\to\infty$ it tends to $1$, thus:
$$\lim_{x\to\infty}\frac{\cos(\frac{1}{c})}{\sin^2(\frac{1}{c}) \cdot x^2}(\psi(x)-\phi(x))=e-1$$
since $(1+\frac{1}{x})^x\to e$ and $\sqrt[x]{x}\to1$
