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If we assume that the restrictions put on simplified forms of expressions to prevent evaluation at points undefined in the original unsimplified form are important why do we drop them when dealing with limits? For example, consider the following when trying to find the derivative of $f= x^2%$:

$$\begin{align*}\lim_{h→0} \frac{f(x + h) - f(x)}{h} &=\lim_{h→0} \frac{(x+h)^2 - x^2}{h}\\ &=\lim_{h→0} \frac{x^2 + 2xh + h^2 - x^2}{h}\\ &=\lim_{h→0} \frac{h(2x + h)}{h} \end{align*}$$

All following simplified forms should have the restriction $h ≠ 0$ since the original form was undefined at this point.

$$\lim_{h→0} {2x + h}, h ≠ 0$$

However to calculate the derivative, the h is valued at $0$ leading to the derivative:

$$f'(x) = 2x$$

How can the equation be simplified by assuming the $h$ is $0$ when there is a restriction on it? Why is that when simplifying expressions we have to restrict the simplified forms to prevent evaluation at points undefined on the original expression, but this concept is completely ignored when dealing with limits?

Arturo Magidin
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  • How do you define $\lim_{h\to0}2x+h$? – sai Jul 11 '12 at 02:35
  • The value of the expression as the value of $h$ approaches $0$ –  Jul 11 '12 at 02:35
  • And wouldn't you agree that this value is $2x$? At no point did we need to set $h=0$. – sai Jul 11 '12 at 02:36
  • $\displaystyle \lim_{h \to 0} (2x+h)$ is not the same as evaluating $2x+h$ at $h=0$. –  Jul 11 '12 at 02:39
  • Exactly...which equals to actually substitute $,h=0,$ since the expression $,2x+h,$ is a linear polynomial in $,h,$ and thus continuous everywhere...This was not so before, when the cancellation was done. Remember, when evaluating the limit of an expression when $,x\to x_0,$, we shall and we must take $,x\neq x_0,$...very close to it, but not actually equal to it...unless we have continuity of the function at $,x_0,$ – DonAntonio Jul 11 '12 at 02:40
  • @sai I see. But the above conclusion suggests that $2x + h$ approaches $2x$ when $h$ approaches zero, however in reality $2x + h$ approaches an undefined value when $h$ approaches zero. How can the formula approach $2x$ AND an undefined value? –  Jul 11 '12 at 02:40
  • Why do you say that $2x+h$ approaches an undefined value? Note that for $\lim_{n\to\infty}x_n=x$, we do not require that at some point $x_n$ equals $x$, but only that it gets arbitrarily close to $x$. For example, $\lim_{n\to\infty}1/n=0$, but the sequence is always some positive distance away from 0. – sai Jul 11 '12 at 02:44
  • @sai But the only way to solve a limit is by subbing in $x$ (the undefined value). You might not be defining the value but when trying to find what the value is you are doing something that is undefined aren't you? You cannot solve a limit without subbing in the value it is approaching right (which is an undefined action)? –  Jul 11 '12 at 02:46
  • You are claiming that one has to substitute the limiting value to evaluate limits? Then consider the well known limit $\lim_{n\to\infty}(1+x/n)^n=e^x$. – sai Jul 11 '12 at 02:48
  • @Riddler, I think you're confusing things: for example, $$\lim_{n\to\infty}\frac{1}{n}=0$$is an exact value of the limit. This does NOT mean that $\frac{1}{n}=0,$ for some $,n\in\Bbb N,$, of course....these are two very different things, and I think that here is where your confusion appears. – DonAntonio Jul 11 '12 at 02:51
  • @sai Your right, I take back my statement. However, for this specific statement you would have to sub in $h = 0$ wouldn't you? How else would you solve this specific limit. –  Jul 11 '12 at 02:55
  • @DonAntonio Thanks that helped! I understand what your saying, but in examples such as the one in my question don' t you have to do undefined actions to arrive at the exact value of the limit? –  Jul 11 '12 at 02:56
  • @Riddler please see my answer. – sai Jul 11 '12 at 03:03
  • Whenever we compute a limit as $x\to a$, the one point we never care about is $x=a$. We are asking what happens near $a$, never at $a$. The restriction $x\neq a$ (or in your case, $h\neq 0$) is implicit in the very act of writing $\lim\limits_{h\to 0}g(h)$. – Arturo Magidin Jul 11 '12 at 03:08

3 Answers3

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In a sense, you are repeating the old criticism of Bishop Berkeley on infinitesimals, which were "sometimes not equal to $0$, and sometimes equal to $0$".

What you need to remember is that the expression $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ represents the unique quantity (if it exists) that the expression $\dfrac{f(x+h)-f(x)}{h}$ approaches as $h$ approaches $0$, but without $h$ being equal to $0$. Whenever we take a limit, we are asking how the quantity is behaving as we approach $0$, but without actually being $0$.

Because we are never actually at $0$, the simplification is valid, and so the computation turns on asking: what happens to the quantity $2x+h$ as $h$ approaches $0$?

The answer is that, the closer $h$ gets to $0$, the closer that $2x+h$ gets to $2x$. We can make $2x+h$ as close to $2x$ as we want, provided that $h$ is close enough to $0$, without being equal to $0$.

We are not actually evaluating at $0$ (well, we kind of are, see below, but not really); we are just finding out what happens to $2x+h$ as $h$ gets closer and closer and closer to $0$. So we are not "simplifying" the way we did before, we are now evaluating the limit, by determining what happens to $2x+h$ as $h$ approaches $0$.

(Now, in a sense we are evaluating, for the following reason: the function $g(h) = 2x+h$, where $x$ is fixed, is continuous and defined everywhere. One of the properties of continuous functions (in fact, the defining property of being continuous) is that $g(t)$ is continuous at $t=a$ if and only if $g$ is defined at $a$, and $$\lim_{t\to a}g(t) = g(a).$$ That is, if and only if the value that the function approaches as the variable approaches $a$ is precisely the value of the function at $a$: there are no jumps, no breaks, and no holes in the graph at $t=a$. But we are not "simplifying" by "plugging in $a$", we are actually computing the limit, and finding that the limit "happens" to equal $g(a)$.

This cannot be done with the original function $\dfrac{f(x+h)-f(x)}{h}$ because, as you note, it is not defined at $h=0$. But there is a result about limits which is very important:

If $f(t)$ and $g(t)$ have the exact same value at every $t$, except perhaps at $t=a$, then $$\lim_{t\to a}f(t) = \lim_{t\to a}g(t).$$

the reason being that the limit does not actually care about the value at $a$, it only cares about the values near $a$.

This is what we are using to do the first simplification: the functions of variable $h$ given by: $$\frac{(x+h)^2-x^2}{h}\qquad\text{and}\qquad 2x+h$$ are equal everywhere except at $h=0$. They are not equal at $h=0$ because the first one is not defined at $h=0$, but the second one is. So we know that $$\lim_{h\to 0}\frac{(x+h)^2 - x^2}{h} = \lim_{h\to 0}(2x+h).$$ And now we can focus on that second limit. This is a new limit of a new function; we know the answer will be the same as the limit we care about, but we are dealing with a new function now. This function, $g(h) = 2x+h$, is continuous at $h=0$, so we know that the limit will equal $g(0)=2x$. Since this new limit is equal to $2x$, and the old limit is equal to the new limit, the old limit is also equal to $2x$. We didn't both take $h\neq 0$ and $h=0$ anywhere. We always assumed $h\neq 0,$ and then in the final step used continuity to deduce that the value of the limit happens to be the same as the value of the function $g(h) = 2x+h$ at $h=0$. )

Cameron Buie
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Arturo Magidin
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Formally, we say $\lim_{x\to a}f(x)=f(a)$ if $\forall \epsilon>0, \exists \delta>0$ such that $|x-a|<\delta\Rightarrow |f(x)-f(a)|<\epsilon$ (except for $x=a$, in case the function is not continuous at $x=a$) or for sequences $\lim_{n\to\infty}a_n=a$ if $\forall \epsilon>0, \exists N\in\mathbb{N}$ such that $|a_n-a|<\epsilon, \forall n>N$. Note that in both definitions, we allow for $\epsilon$ tolerance, which is positive - but may be arbitrarily small.

In your example, $\lim_{h\to0}2x+h=2x$ since $(2x+h)-2x=h$, which can be made arbitrarily small. Alternatively, you can think of a line with slope = 2. What the limit is saying is that as the vertical shift of the line $h$ approaches 0, the line will get arbitrarily close to the slope 2 line through the origin.

sai
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For limits, we aren't (in general) simply evaluating at a particular point. Intuitively, when we're dealing with real-valued functions on subsets of the reals, we talk about the graph of a function getting as close as we like to a point if we make the $x$-coordinate get sufficiently close to the point's $x$-coordinate.

Rigorously, we use the following definitions:

(i) Suppose $E\subseteq\Bbb R$, $x_0\in\Bbb R$. We say that $x_0$ is a limit point of $E$ if and only if for every $\varepsilon>0$ there is some $x\in E$ such that $0<|x-x_0|<\varepsilon$. (That is, there are points of $E$ that are as close to $x_0$ as we like, but still distinct from $x_0$.)

(ii) Suppose $E\subseteq\Bbb R$, $x_0\in\Bbb R$ a limit point of $E$, $L\in\Bbb R$, and $f:E\to\Bbb R$ is some function. We say that $$L=\lim_{x\to x_0}f(x)$$ if and only if for every $\varepsilon>0$ there exists $\delta>0$ such that $|f(x)-L|<\varepsilon$ whenever $x\in E$ and $0<|x-x_0|<\delta$. (That is, $L=\lim_{x\to x_0}f(x)$ means we can make $f(x)$ as close as we like to $L$ simply by choosing $x\in E$ sufficiently close to $x_0.$ Note that $x_0$ need not be an element of $E.$)

For your example, we're dealing with an $h$ limit (so replace the $x$ terms above with $h$ terms). Letting $\delta=\varepsilon$, we find that for $0<|h-0|<\delta$, we have $$|(2x+h)-2x|=|h|=|h-0|<\delta=\varepsilon.$$

Cameron Buie
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