In calculus we are taught that the derivative of function $y$ with respect to $x$ is defined as "the quantity which $\dfrac{\bigtriangleup{y}}{\bigtriangleup{x}}$ tends to when $\bigtriangleup{x}$ tends to zero"
Can we also define it as "the ratio of $\dfrac{\bigtriangleup{y}}{\bigtriangleup{x}}$ when $\bigtriangleup{x}$ is a quantity very close to zero"?
Let $y = x^2$. Then $\dfrac{dy}{dx} = 2x$ for the usual reasons
$\lim_{\Delta x \rightarrow 0} \dfrac {\Delta y}{\Delta x} = \lim \dfrac {[x^2 + 2x\Delta x + (\Delta x)^2] -[x^2]}{\Delta x} = \lim (2x + \Delta x)= 2x$.
But if we chose to say that the derivative is a value where $\Delta x$ is a number "very near zero" (I'm going to ignore the difficulty of how one would define such an ambiguous expression) so that $\Delta x = \delta >0$ but $\delta$ is "small".
Then $\dfrac {\Delta y}{\Delta x} = \dfrac {(x+\delta)^2 - x^2}{\delta} \dfrac {x^2 + 2x\delta + \delta^2 }{\delta} = 2x + \delta \ne 2x$.
So, no, that is a different answer. You can't say $\Delta x$ isn't zero at the beginning and then say "well $\Delta x$ very close to $0$ so we can ignore it" in the end.