Isn't $\lim_{h \to 0} \frac{c-c}{h}$ indeterminate, as $\lim_{h \to 0} \frac{c-c}{h} = \frac{0}{0}$?

Question

I have a question that asks me to differentiate $f(x) = e^5$. This looks like differentiating a constant. So the answer is 0.

But I'm confused about the proof withthe definition of a deriative:

Don't we have an indeterminate limit since $\lim_{h \to 0} \frac{c-c}{h} = \frac{0}{0}$?

Or is the right way to think about this is that the numerator is 0 but the denominator is never really quite 0 and so the limit as a whole is 0?

Answer 1

No, $c-c = 0$ always, even prior to letting $h$ go to zero. So you have $$ \lim_{h \rightarrow 0} \frac{c-c}{h} = \lim_{h \rightarrow 0} \frac{0}{h} = \lim_{h \rightarrow 0} 0 = 0 \text{.}$$

That is, you take the limit of the constant expression zero.

Don't try to do "everything at once". You must evaluate inner nested expressions before starting to evaluate their enclosures. In this case, evaluate the fraction as completely as possible, then take the limit. Correctly evaluating nested expressions from the inside to the outside will be very important when you have nested limits.

Answer 2

The form $0/0$ is indeed indeterminate, so you cannot compute

$$ \lim_{h \to 0} \frac{c-c}{h} $$

simply by substituting in the limits of the numerator and denominator. However, that's not what the proof did. The proof continued on to make an algebraic simplification

$$ \frac{c-c}{h} = 0 \qquad \qquad (h \neq 0) $$ and thus

$$ \lim_{h \to 0} \frac{c-c}{h} = \lim_{h \to 0} 0 $$

and this limit is easy to compute.

Answer 3

You have a misunderstanding about $\frac00$ and $\frac0h$

When we have $\frac0x$ we get $0$ unless $x=0$, but the $\lim_{x\to0}\frac0x$ is not $\frac00$. You can't say $\lim_{x\to0}\frac0x=\frac0{\lim_{x\to0}x}$(the function $\frac ax$ is discontinuous at $x=0$).

Before taking the limit we have to calculate $\frac0x=0$, so we get $\lim_{x\to0}0=0$