To compute $$\lim_{x \to 0}{(1+x)^{1 \over x}-(1+2x)^{1 \over 2x} \over \sin x}$$
and we know that $e^{{1 \over x}\ln(1+x)} = {(1+x)^{1 \over x}}$ and $\lim_{x \to 0}{{1 \over x}\ln(1+x)}=0$
Why can't I compute this as following:
$$\lim_{x \to 0}{(e^{{1 \over x}\ln(1+x)}-1)-(e^{{1 \over 2x}\ln(1+2x)}-1) \over x}$$ $$\lim_{x \to 0}{{{1 \over x}\ln(1+x)-{1 \over 2x}\ln(1+2x)}\over x}$$
$\lim_{x\to 0} e^x-1=x$ is meaningless.
If you mean $(\lim_{x\to 0} e^x-1)=x$, the left side is a number, while the right side is a variable (function), how are they equal?
If you mean $\lim_{x\to 0} (e^x-1=x)$, then what's the meaning of a limit of an equation?
I guess you mean $\lim_{x\to 0}\frac{e^x-1}{x}=1$, but you cannot use this to replace $e^x-1$ with $x$ to calculate other limits, because there is no rule allowing you to do so.