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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?
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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phi
 one year ago
Best ResponseYou've already chosen the best response.1see 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 \[ \lim_{x\rightarrow a}x\cdot g(x)=x\cdot \lim_{x\rightarrow a} g(x) \text{ WRONG} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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?

phi
 one year ago
Best ResponseYou've already chosen the best response.1you 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My 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.

phi
 one year ago
Best ResponseYou've already chosen the best response.1let'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)

phi
 one year ago
Best ResponseYou've already chosen the best response.1and to be clearer \[ \lim_{x\rightarrow a}x\cdot f(x)\] means take the limit of the product x*f(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, 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.

phi
 one year ago
Best ResponseYou've already chosen the best response.1I 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, \[ 2x +1 = x4\] 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In 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?

phi
 one year ago
Best ResponseYou've already chosen the best response.1everything 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 welldefined. if you have time, see http://ocw.mit.edu/resources/res18006calculusrevisitedsinglevariablecalculusfall2010/partisetsfunctionsandlimits/lecture4derivativesandlimits/ and the next lecture http://ocw.mit.edu/resources/res18006calculusrevisitedsinglevariablecalculusfall2010/partisetsfunctionsandlimits/lecture5amorerigorousapproachtolimits/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is a different case in which the scopes are well behaved. The odd case is as was described.
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