$dX_t/X_t=\mu\, dt+\sigma \, dZ_t$, does this notation make sense? I understand that the notation

$$dX_t=\mu X_t \,dt + \sigma X_t \,dZ_t,$$ 

where $Z_t$ is Brownian Motion, is a shortcut to

$$X_t-X_0=\int_0^t\mu X_s \, ds+\int_0^t \sigma X_s \, dZ_s, \tag{*}$$

which has a precise meaning, since the Ito integral is defined. 

But I don't get it when people write:

$$\frac{dX_t}{X_t}=\mu \, dt + \sigma \, dZ_t,$$ 

and later go on defining $\frac{dX_t}{X_t}$ as a variable. What does $\frac{dX_t}{X_t}$ even mean? To me, it looks like taking $X_s$ out of the integral in $(*)$, which does not make any sense.

For example, if $X_t$ is the price of an asset, $dX_t/X_t$ is sometimes defined as the rate of return the asset, but my point is that it is not even a mathematically defined object.

Am I missing something, or this is sloppy?
$$\frac{dx_t}{x_t}=\mu+\sigma dz_t\\$$let $f(x)=ln(x)$ apply ito formula to find $df(x)$ $$df=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial x}dx+\frac{1}{2}\frac{\partial^2f }{\partial x^2}\\df=0dt+\frac{1}{x}dx+\frac{1}{2}(\frac{-1}{x^2})dx^2\\df=0dt+\frac{1}{x}dx+\frac{1}{2}(\frac{-1}{x^2})dx^2\\$$now put $dx_t=\mu x_tdt+\sigma x_t dz_t$
$$df=0dt+\frac{1}{x_t}(\mu x_tdt+\sigma x_t dz_t)+\frac{1}{2}(\frac{-1}{x_t^2})(\mu x_tdt+\sigma x_t dz_t)^2\\$$
$$df=0dt+\frac{1}{x_t}(\mu x_tdt+\sigma x_t dz_t)+\frac{1}{2}(\frac{-1}{x_t^2})(\mu^2 x_t^2(dt)^2+\sigma^2 x^2_t (dz_t)^2+2\mu \sigma x^2_tdtdz_t)^2\\$$
apply $dt.dt \to 0\\dt.dz_t \to 0 \\dz_t.dz_t \to dt$
so 
$$df=0dt+\mu dt +\sigma dz_t-\frac{1}{2}\sigma^2 dt$$apply integral for both sides
$$\int_{0}^{t}d(lnx_s)=\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s\\ \ln(x_t)-\ln(x_0)=\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s\\\frac{x_t}{x_0}=e^{\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s}\\x_t=x_0.e^{\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s}$$
$$\frac{dx_t}{x_t}=\mu+\sigma dz_t\\$$let $f(x)=ln(x)$ apply ito formula to find $df(x)$ $$df=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial x}dx+\frac{1}{2}\frac{\partial^2f }{\partial x^2}\\df=0dt+\frac{1}{x}dx+\frac{1}{2}(\frac{-1}{x^2})dx^2\\df=0dt+\frac{1}{x}dx+\frac{1}{2}(\frac{-1}{x^2})dx^2\\$$now put $dx_t=\mu x_tdt+\sigma x_t dz_t$
$$df=0dt+\frac{1}{x_t}(\mu x_tdt+\sigma x_t dz_t)+\frac{1}{2}(\frac{-1}{x_t^2})(\mu x_tdt+\sigma x_t dz_t)^2\\$$
$$df=0dt+\frac{1}{x_t}(\mu x_tdt+\sigma x_t dz_t)+\frac{1}{2}(\frac{-1}{x_t^2})(\mu^2 x_t^2(dt)^2+\sigma^2 x^2_t (dz_t)^2+2\mu \sigma x^2_tdtdz_t)^2\\$$
apply $dt.dt \to 0\\dt.dz_t \to 0 \\dz_t.dz_t \to dt$
so 
$$df=0dt+\mu dt +\sigma dz_t-\frac{1}{2}\sigma^2 dt$$apply integral for both sides
$$\int_{0}^{t}d(lnx_s)=\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s\\ \ln(x_t)-\ln(x_0)=\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s\\\frac{x_t}{x_0}=e^{\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s}\\x_t=x_0.e^{\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s}$$