Recall the definition of a derivative: f'(x)=f(x+∆x) - f(x) / ∆x Explain why, if the denominator ∆x is going to zero, the fraction f(x+∆x)−f(x) doesn’t necessarily ∆x go to infinity. (Hint: The numerator is also going to zero... so what’s the fraction doing?)

Step-by-step explanation:

Given the definition of a derivative: f'(x)=f(x+∆x) - f(x) / ∆x, if we take the limit as ∆x tends to zero, the derivative will become;

f'(x) = f(x+∆x) - f(x) / ∆x

f'(x) = f(x+0) - f(x) / 0

f'(x) = f(x) - f(x) / 0

Since f(x)-f(x) = 0, the derivative will become;

f'(x) = 0/0 (indeterminate)

As we can see, the derivative doesn't turn to infinity because the numerator of function also tends to zero as ∆x--> 0.

Instead of the fraction to tend to infinity, it becomes an indeterminate function and an indeterminate function can be solved further by applying l'hospital rule on the resulting function and re-substituting ∆x. There is possibility that the equation will give a finite value afterwards depending on the function.