## anonymous one year ago 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?

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1. phi

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 $\lim_{x\rightarrow a}x\cdot g(x)=x\cdot \lim_{x\rightarrow a} g(x) \text{ WRONG}$

2. anonymous

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?

3. phi

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?

4. anonymous

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.

5. phi

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)

6. phi

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

7. anonymous

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.

8. phi

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.

9. anonymous

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?

10. phi

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/

11. anonymous

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