Since derivatives are continuous already, we can put h = 0.
Hello,
for many years i have been doing derivatives, and its definition: lim f(x+h)-f(x) / h always seemed quite obvious to me, but then i started to think about it...
h->0
when we do derivatives through its definition, we usually arrive to something like (h(x+h))/h and we remove the h on the numerator and denominator, leaving x+h, when we replace h by 0 and get to x....but the question is, wont we be doing a violation when we place h by 0? If we condsidered the previous formula, replacing 0 by 0 would cause a 0/0 indetermination, which arises from the fact that the removal rule only applices if h!=0, so....what is that last result we get? The result of ignoring the fact that h cannot be 0 but at the same time assume it is 0?
Thanks in advance
What make you say that. That is false.
If a function has a derivative at then the function is continuous at
We don't about the derivative at .
When evaluating we never just let .
That is a fundamental principle of limits. We must look for continuity.
When considering the difference quotient we hope that the divides out.
That can only be if
WE DON'T.
We simply see what happens for values of h nearly zero.
Lazy teachers allow replacement. But it really is not correct.
If the difference quotient reduces to a continuous function of h, then it works.
It works, but technically not correct.
Because a derivative is a limit, we must follow the formal definition of limit.
In the definition of it is required that