Question

There's this: c * lim_(x>a) f(x) = lim_(x>a) c * f(x)

Why isn't there also this,

x * lim_(x>a) f(x) = lim_(x>a) x * f(x)?

Not for any function of x, but just x?

Answer 1

see http://planetmath.org/ProofOfLimitRuleOfProduct
there is this
\[ \lim_{x\rightarrow a} f(x)\cdot g(x)= \lim_{x\rightarrow a} f(x)\cdot \lim_{x\rightarrow a} g(x) \]

one valid f(x) is f(x)= x
thus we can say
\[ \lim_{x\rightarrow a}x\cdot g(x)= \lim_{x\rightarrow a} x\cdot \lim_{x\rightarrow a} g(x) \]
and of course the limit as x goes to a of x is just a
\[ \lim_{x\rightarrow a}x\cdot g(x)=a\cdot \lim_{x\rightarrow a} g(x) \]
notice we cannot get
\[ \lim_{x\rightarrow a}x\cdot g(x)=x\cdot \lim_{x\rightarrow a} g(x) \text{ WRONG} \]

Answer 2

I see that we do not get
\[\lim_{x \rightarrow a} [x \times f(x)] = x \times \lim_{x \rightarrow a} f(x)\]
in that way, but not that we cannot get it in another way. Can you elaborate please?

Answer 3

you can have the expression
\[ x\cdot \lim_{x\rightarrow a} f(x) \]
and you can have the expression
\[ \lim_{x\rightarrow a} x\cdot f(x) \]
but they are not equal expressions.

I may not be understanding your question. Perhaps if you ask it a different way?

Answer 4

My question is whether the two expressions are equal? And, if so, why so? And, if not, why not?

I understand that x and x + 1 are not equal expressions. Because, I think, there is no value of x that will make x = x + 1 true. Is it the same with

x * lim_(x>a) f(x)

and

lim_(x>a) x * f(x)?

If it is, I just don't see how it works. A sentence like "They are not equal, because...." would be helpful.

Answer 5

let's say the limit has some value:
\[\lim_{x\rightarrow a} f(x)= L\]
then
\[ x\cdot \lim_{x\rightarrow a} f(x)= xL\]
on the other hand
\[ \lim_{x\rightarrow a}x\cdot f(x)= \lim_{x\rightarrow a}x\cdot \lim_{x\rightarrow a} f(x)
=aL
\]
in general (i.e. for all values of x)
xL is not equal to aL
(though obviously for *some* x values, (i.e. when x=a) , the expressions are equal)

Answer 6

and to be clearer
\[ \lim_{x\rightarrow a}x\cdot f(x)\]
means take the limit of the product x*f(x)

Answer 7

Ok, then,

x * lim_(x>a) f(x) = lim_(x>a) x * f(x)

is not like x = x + 1, because there are no values of x that make the second true, but there is one value of x that makes the first true. Moreover, if the first, then x = a. What is the problem with this? Perhaps if you describe a similar situation, it might make the point clearer.

Answer 8

I guess it is a matter of communication
if we have two functions f(x) and g(x)
we can ask (at least) these questions:
is f(x) the same as g(x)
is f(x) = g(x) for some x

when we "simplify" expressions, we mean the first question
for example
\[ x\cdot x = x^2 \]
and this is a different question from, for example,
\[ 2x +1 = x-4\]
which is only true for some specific value of x

anyway,
the limit question you are asking are more like the first interpretation. Or , at least, that is how people treat them.

Answer 9

In sum, I observed that

c lim (x > a) f(x) = lim (x >a) c f(x).

Then I asked why not

x lim (x > a) f(x) = lim (x >a) x f(x)? (Call this equation E.)

In short, it seems that the answer begins with the further observation that my question is essentially whether E is an identity, that is, whether E is true for all values of x. Since E is only true for one value of x, E is not an identity. In short, that is why not E. I understand this answer.

But, there is something odd about E. From the point of view of free and bound variables, one occurrence of x (marked as [x]) is not bound by the limit operator and is either free or bound by some other operator or quantifier. The other occurrence of x (marked as [[x]]) is either free or bound by another implicit quantifier:

[x] lim (x > a) f(x) = lim (x >a) [[x]] f(x)

I don't know whether this matters, but it would seem to make E ill formed rather than false?

Answer 10

everything you said sounds good except for that
\[ \lim_{x\rightarrow a}x\cdot f(x)= \lim_{x\rightarrow a}x\cdot \lim_{x\rightarrow a} f(x)= a \cdot \lim_{x\rightarrow a} f(x)
\]
is well-defined.
if you have time, see
http://ocw.mit.edu/resources/res-18-006-calculus-revisited-single-variable-calculus-fall-2010/part-i-sets-functions-and-limits/lecture-4-derivatives-and-limits/
and the next lecture
http://ocw.mit.edu/resources/res-18-006-calculus-revisited-single-variable-calculus-fall-2010/part-i-sets-functions-and-limits/lecture-5-a-more-rigorous-approach-to-limits/

Answer 11

That is a different case in which the scopes are well behaved. The odd case is as was described.