UPD1: By $g(x)$ I mean not any expression, but some non constant expression ($\lim_{x-1\to0} f(x-1)$, $\lim_{x^3\to0} f(x^3)$, etc.).
UPD2: $g(x)$ should have the property that $\exists a\ \lim_{x\to a} g(x)=0$.
I'm currently reading Spivak's Calculus, and the book states that $x$ is irrelevant in the notation $\lim_{x\to a} f(x)=l$, the only significant things being $f$, $a$ and $l$. So I draw a conclusion that $\lim_{x\to0} f(x)=\lim_{g(x)\to0} f(g(x))$ (provided that $\lim_{x\to0} f(x)$ exists).
To prove it formally, I suppose that the first limit exists, $\lim_{x\to0} f(x)=l$,
$$\forall\epsilon>0,\ \exists\delta,\ \forall x,\ 0<|x|<\delta\implies |f(x)-l|<\epsilon.$$
Which also means (because of the "given" implication above):
$$\forall\epsilon>0,\ \exists\delta,\ \forall x,\ 0<|g(x)|<\delta\implies |f(g(x))-l|<\epsilon$$ (provided that $|g(x)|$ is defined on $(0,\delta)$).
Is that correct? If it is, then the limit definition could also be stated as:
$\lim_{x\to a} f(x) = l \Leftrightarrow \forall \epsilon>0 \ \exists\delta\ 0<|g(x)-a|<\delta \implies |f(g(x))-l| < \epsilon)$, with $g(x)$ being some expression involving $x$, defined on $(0,\delta)$ ?
Best Answer
The correct term is dummy and $x$ is a dummy variable in the notation $$\lim_{x\to a} f(x)=l$$ or in $$I=\int_{a} ^{b} f(x) \, dx$$ This is because the definition of limit deals with the function $f$, the point $a$ under consideration and the proposed limit $l$.
You can change the variable $x$ which also occurs in definition of limit to some other symbol say $t$ and the definition remains valid for $\lim_{x\to a} f(x) =l$. Instead if you change the symbol $l$ in definition to $m$ the definition is not valid for $\lim_{x\to a} f(x) =l$ but instead it now works for $\lim_{x\to a} f(x) =m$. This way the usage of variable $x$ in the definition is very different from that of $f, a, l$.
Consider the following analogous example. Let $$A=\{x\mid x \text{ is a prime number} \} $$ then we can also write $$A=\{p\mid p\text{ is a prime number}\} $$ Here both $x, p$ are dummy variables but $A$ is not.
In general you can't replace a dummy variable with something which is not a dummy variable.
The result which you are trying to write is more properly known as rule of substitution :
Your case is $a= b=0$. Using this rule you can conclude $$\lim_{x\to 0}\frac{\log(1+x)}{x}=1\implies\lim_{x\to 0}\frac{\log(1+\sin x)} {\sin x} =1$$ And you can also note that instead of $\sin x$ you can have any function which tends to $0$ (but does not equal $0$) with $x$ (eg $\cos x - 1$).
If some instructor / examiner is hell bent on showing all steps in detail this is how one would evaluate the limit of $\lim_{x\to 0}\dfrac{\log(1+\sin x)} {\sin x} $.
Let us put $t=\sin x$ so that $t\to 0$ as $x\to 0$ and the desired limit is reduced to $$\lim_{t\to 0}\frac{\log (1+t)}{t}$$ which is a standard limit in textbook with value $1$. The substitution $t=\sin x$ is justified because $\sin x\neq 0$ as $x\to 0$.
If there is no need for such details you can directly write $$\lim_{x\to 0}\frac{\log(1+\sin x)} {\sin x} =1$$ You may also observe that we don't use the notation $$\lim_{\sin x \to 0}\frac{\log(1+\sin x)} {\sin x} =1$$ like you are trying to do.