anonymous
  • anonymous
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?
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
phi
  • 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 \[ \lim_{x\rightarrow a}x\cdot g(x)=x\cdot \lim_{x\rightarrow a} g(x) \text{ WRONG} \]
anonymous
  • 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?
phi
  • 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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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.
phi
  • 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)
phi
  • phi
and to be clearer \[ \lim_{x\rightarrow a}x\cdot f(x)\] means take the limit of the product x*f(x)
anonymous
  • 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.
phi
  • 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.
anonymous
  • 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?
phi
  • 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/
anonymous
  • anonymous
That is a different case in which the scopes are well behaved. The odd case is as was described.

Looking for something else?

Not the answer you are looking for? Search for more explanations.