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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 $\lim_{\delta x\to0} \frac{\delta x}{\delta x}$ ?

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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.

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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 $1$.

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