Solve $\lim_{x\to0^+} \frac{(1+\sin x)^{\frac1x} - \exp(-\frac x2)}{(\tan x )^\alpha}$

Question

Is this limit solved correctly?

$$\lim_{x\to0^+} \frac{(1+\sin x)^{\frac1x} - \exp(-\frac x2)}{(\tan x )^\alpha}$$

$$\lim_{x\to0^+} \frac{\exp\left(\frac1x\ln(1+\sin x)\right) - \exp(-\frac x2)}{(\tan x )^\alpha}$$

$$\lim_{x\to0^+} \frac{\exp\left(\frac1x\ln(1+x+o(x^2)\right) - \exp(-\frac x2)}{(\tan x )^\alpha}$$

$$\lim_{x\to0^+} \frac{\exp\left(\frac1x\left(\left[x+o(x^2)\right]-\frac12\left[x+o(x^2)\right]^2 + o(x^2)\right)\right) - \exp(-\frac x2)}{(\tan x )^\alpha}$$

$$\lim_{x\to0^+} \frac{\exp\left(\frac1x\left[x-\frac{x^2}{2} + o(x^2)\right]\right) - \exp(-\frac x2)}{(\tan x )^\alpha}$$

$$\lim_{x\to0^+} \frac{\exp\left(1-\frac{x}{2} + o(x)\right) - \exp(-\frac x2)}{(\tan x )^\alpha}$$

$$\lim_{x\to0^+} \frac{\exp(1)\cdot\exp\left(-\frac{x}{2}\right)\cdot\exp(o(x)) - \exp(-\frac x2)}{(\tan x )^\alpha}$$

$$\lim_{x\to0^+} \frac{\exp\left(-\frac{x}{2}\right)\cdot\left(e\cdot\exp(o(x)) - 1\right)}{x^\alpha} = \frac{1\cdot(e\cdot1 -1)}{\lim_{x\to0^+}x^\alpha} = \begin{cases}\frac{e-1}{0^+} = +\infty && \ \text{if } \alpha >0\\\frac{e-1}{1} = e-1 &&\ \text{if } \alpha =0\\\frac{e-1}{+\infty} = 0 &&\ \text{if } \alpha <0\end{cases}$$

EDIT:

A first-order approximation is enough:

$$\lim_{x\to0^+} \frac{\exp\left(\frac1x\ln(1+x+o(x^2)\right) - \exp(-\frac x2)}{(\tan x )^\alpha}$$

$$\lim_{x\to0^+} \frac{\exp\left(\frac1x\left[x+o(x^2)+o(x)\right]\right) - \exp(-\frac x2)}{(\tan x )^\alpha}$$

$$\lim_{x\to0^+} \frac{\exp\left(1+o(1)\right) - \exp(-\frac x2)}{(\tan x )^\alpha} = \frac{e-1}{\lim_{x\to0^+}x^\alpha} = \begin{cases}\frac{e-1}{0^+} = +\infty && \ \text{if } \alpha >0\\\frac{e-1}{1} = e-1 &&\ \text{if } \alpha =0\\\frac{e-1}{+\infty} = 0 &&\ \text{if } \alpha <0\end{cases}$$