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.
 A: 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 $\sqrt[x]{x}\to1$
A: By the limit chain rule, we can distribute the limits into the sins:
$$\lim_{x\to\infty}[\frac{1}{\sin(\frac{1}{x+\lim_{x\to\infty}\psi(x)})}-\frac{1}{\sin(\frac{1}{x+\lim_{x\to\infty}\phi(x)})}]$$
and calculate $\lim_{x\to\infty}(1+\frac{1}{x})^x$ and $\lim_{x\to\infty}\sqrt[x]{x}$ first. The former is a standard limit and evaluates to $e$ the latter is done easily by L'Hopital's rule:
$$\lim_{x\to\infty}\sqrt[x]{x} = \lim_{x\to\infty}e^{ln(x^{\frac{1}{x}})}$$
$$= \lim_{x\to\infty}e^{\frac{ln(x)}{x}}$$
$$= e^{\lim_{x\to\infty}\frac{ln(x)}{x}}$$
$$= e^{0}$$
$$= 1$$
So our limit becomes:
$$\lim_{x\to\infty}[\frac{1}{\sin(\frac{1}{x+e})}-\frac{1}{\sin(\frac{1}{x+1})}]$$
The functions $\frac{1}{\sin(\frac{1}{x+e})}$ and $\frac{1}{\sin(\frac{1}{x+1})}$ can be approximated by $x+e$ and $x+1$ respectively arbitrarily well when x tends towards infinity. The proof is as follows:
The taylor series of $\csc(x)$ is of the form:
$$\csc(x)=\sum_{i=0}^\infty (c_ix^{2i-1})$$
Where $c_i$ is a series of strictly decreasing rational constants and $c_0 = 1$. Substituting $c=\frac{1}{u}=u^{-1}$ we have:
$$\frac{1}{\sin(x)} = \sum_{i=0}^\infty (c_ix^{1-2i}) = \frac{1}{x} + c_1x + c_2x^3 + \, ...$$
$$\frac{1}{\sin(u^{-1})} = \sum_{i=0}^\infty (c_iu^{1-2i}) = u + c_1u^{-1} + c_2u^{-3} + \, ...$$
Taking the limit as u tends towards infinity, everything except for the first term drop out because $1-2i < 0$ for i > 0:
$$\lim_{u\to\infty}c_iu^{1-2i} = 0; i > 0$$
Therefore, rewriting gives:
$$\lim_{u\to\infty}\frac{1}{\sin(u^{-1})} = \lim_{u\to\infty}\sum_{i=0}^\infty (c_iu^{1-2i}) = \lim_{u\to\infty}(u + c_1u^{-1} + c_2u^{-3} + \, ...)=\lim_{u\to\infty}(u)$$
If we substitute $x+e$ and $x+1$ for $u$ then we have:
$$\frac{1}{\sin(\frac{1}{x+e})} \approx x + e$$
$$\frac{1}{\sin(\frac{1}{x+1})} \approx x + 1$$
$$\lim_{x\to\infty}[\frac{1}{\sin(\frac{1}{x+e})}-\frac{1}{\sin(\frac{1}{x+1})}] = \lim_{x\to\infty}[(x+e)-(x+1)] = \lim_{x\to\infty}[e-1] = e-1$$
