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I'm taking Calculus 1 in college, and one aspect about the limit definition of a derivative really confuses me and I wasn't able to find an answer on the site that made complete sense to me.

Concerning the $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

definition of the derivative, why does factoring out the h give a different result? If h is approaching $0$, shouldn't plugging in $0$ for $h$ at any time give an undefined answer?

This probably seems pretty elementary, but I'm just confused as to why the same equation, in essence, can yield a different outcome.

Lucas
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wagboi
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  • Just to clarify : You are saying that there are two ways of solving the limit : 1 : plugging in $0$ for $h$ gives an undefined answer, and 2 : factoring out the $h$, various things cancel out and you get a number for an answer, and you don't understand why these two approaches are different. Is that right? – Sarvesh Ravichandran Iyer Sep 14 '18 at 05:45
  • Yes; exactly. I've been taught how to find the limit, and I can go through the motions, but I don't exactly understand what I'm doing conceptually. – wagboi Sep 14 '18 at 05:59
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    The idea of a limit is something very very different from that of plugging. However in a vast majority of situations it turns out that effect of limit is same as effect of plugging. See https://math.stackexchange.com/a/1822706/72031 for details. – Paramanand Singh Sep 14 '18 at 06:01
  • Plugging in $0$ into $\frac hh$ gives $\frac 00$, but $\frac hh = 1$. So clearly plugging in has problems with the denominator, which we look to eliminate using cancellation with the numerator if it can happen. – Sarvesh Ravichandran Iyer Sep 14 '18 at 06:13

2 Answers2

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I think your confused about what a “limit” really means.

When I say $\lim_{x \to 1} \frac{x^2-1}{x-1}=2$, I mean to say $f(x)=\frac{x^2-1}{x-1}$ gets (arbitrarily) closer to $2$ when $x$ gets closer to $1$ (but is not exactly $1$).

As an example of what I mean, notice $f(.9)=1.9$ and $f(.99)=1.99$ and $f(.999)=1.999$. And similarly if we get closer to $1$ (approach $x=1$) from numbers larger than $1$, we get something similar. $f(1.1)=2.1$, $f(1.01)=2.01$, $f(1.001)=2.0001$.

In fact it seems for $x \neq 1$, that $f(x)=x+1$, in which case I think it’s pretty clear that when $x$ gets closer to $1$, but is not $1$, that $x+1$ gets closer to $2$.

In your specific question $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ (if it exists) is a representation of the value that $\frac{f(x+h)-f(x)}{h}$ gets (arbitrary) closer to as $h$ gets closer to zero (but is not zero).

Ahmed S. Attaalla
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Let's take a specific example; when plugging in $f(x) = x^2$ into $\displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$, we end up with

$$\lim_{h \to 0} \dfrac {h(2x+h)}{h}$$

Now let's say we're trying to find the derivative at $x = 3$ (so that $x$ is constant now) and $h$ is the variable. In other words, let's graph the function $g(h) = \dfrac {h(2 \cdot 3+h)}{h}$. If you do so, it will be a line; however, it will have a hole at $h = 0$. Now, from the graph, we can see the value that $g(0)$ "should have"; the value is $6$. However, algebraically we can't tell that from $\dfrac {h(2 \cdot 3+h)}{h}$ because plugging in $h = 0$ makes it undefined. So to help us see it algebraically, we cancel out the $h$'s and get $6+h$, and also get the answer $6$. If you graph $6+h$, it is exacly the same graph as $\dfrac {h(2 \cdot 3+h)}{h}$, except with the hole filled in.

In summary, canceling the $h$ fills in the hole in the graph and thus allows us to determine the limit algebraically, rahter than having to draw the graph and look visually.

Ovi
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