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?

Limits don't actually care about the value that they approach; they only care about the value that they appear to approach. — Nat, Jan 18 '18 at 00:54

score 29 · Answer 1 · answered Jan 17 '18 at 06:53

29

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.

answered Jan 17 '18 at 06:53

7

A common misunderstanding is considering “indeterminate form” to mean the limit cannot be computed. But of course this is not true and your phrasing is quite good. – egreg Jan 17 '18 at 12:30
Although I would never think of an indeterminate form when I see this limit, by definition of an indeterminate form it is indeed one. +1 – Paramanand Singh Jan 18 '18 at 13:37

score 23 · Accepted Answer · answered Jan 17 '18 at 06:55

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.

score 3 · Answer 3 · edited Jan 18 '18 at 04:38

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 $\lim_{x\to0}0=0$

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

3 Answers3