Why is $\lim_{\delta x\to0} \frac{\delta x}{\delta x} = 1$

Why is $$\lim_{\delta x\to0} \frac{\delta x}{\delta x} = 1$$, considering that both are infinitesimally small but may be different from each other?

Also, if so, why can I not replace $$\frac{\delta f}{\delta x} = \frac{\frac{1}{x + \delta x} - \frac{1}{x}}{\delta x} = 1$$ directly instead of having to reduce it first, since it already amounts to $$\lim_{\delta x\to0} \frac{\delta x}{\delta x}$$ ?

Why is $$\lim_{\delta x\to0} \frac{\delta x}{\delta x} = 1$$, considering that both are infinitesimally small but may be different from each other?

No, they are never different from each other. $$\delta x = \delta x$$. It is the same variable. Furthermore for $$\delta x \neq 0$$ we have $$\frac{\delta x}{\delta x} = 1$$, hence you get $$\lim_{\delta x\to 0} 1 = 1$$.

Also, if so, why can I not replace $$\frac{\delta f}{\delta x} = \frac{\frac{1}{x + \delta x} - \frac{1}{x}}{\delta x} = 1$$ directly instead of having to reduce it first, since it already amounts to $$\lim_{\delta x\to0} \frac{\delta x}{\delta x}$$ ?

It looks like you are considering the derivative of $$f(x)=\frac 1 x$$. Fixing $$x$$ you set $$\delta f = f(x+\delta) - f(x)$$ for every $$\delta x\neq 0$$. Note that $$\delta f$$ and $$\delta x$$ are different things and hence the quotient is not just $$1$$. The derivative is then $$f'(x) = \lim_{\delta x\to 0} \frac{\delta f}{\delta x} = \lim_{\delta x\to 0} \frac{\frac{1}{x+\delta x}-\frac{1}{x}}{\delta x},$$ which is not related to $$\frac{\delta x}{\delta x}$$ at all.

Notice that for $$\frac{\delta x}{\delta x}$$, the numerator and the denominator uses the same symbol. Hence, we have $$\frac{\delta x}{\delta x}=1$$

as long as $$\partial x$$ is not equal to $$0$$.

However, for the second case, $$\frac1{x+\delta x}-\frac1x=\frac{-\partial x}{x(x+\delta x)}$$ doesn't reduces to $$\delta x$$.

Hence $$\lim_{\partial x\to 0}\frac{\frac1{x+\delta x}-\frac1x}{\partial x}=\lim_{\partial x \to 0}\frac{-1}{x(x+\delta x)}=-\frac1{x^2}$$

and we can see that it need not be equal to $$1$$.