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November 19th, 2013, 06:40 PM   #11
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Re: In what situation frac of limits equals to limit of fr

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 Originally Posted by agentredlum What if you do this ? $10^x \= \ e^x \ \cdot \ \frac{10^x}{e^x}$ $\frac{ \lim_{x \to +\infty} \ e^x}{\lim_{x \to +\infty} \ 10^x } \= \ \frac{ \lim_{x \to +\infty} \ e^x}{ \lim_{x \to +\infty} \ $$e^x \cdot \frac{10^x}{e^x }$$ } \ = \ \frac{ \cancel{ \lim_{x \to +\infty} \ e^x}}{ \cancel{ \lim_{x \to +\infty} \ e^x} \ \cdot \ \lim_{x \to +\infty} \ $$\frac {10}{e}$$^x} \ = \ \frac{1}{ + \infty} \ = \ 0$ Since $\frac{ 10}{e} \ > \ 1$
Yeah, this Kingkong physicist did the same thing.

 November 19th, 2013, 07:37 PM #12 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,898 Thanks: 1093 Math Focus: Elementary mathematics and beyond Re: In what situation frac of limits equals to limit of fr $\lim_{x\to a}\,\frac{f(x)}{g(x)}\,=\,\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$ but what are the conditions on this?
November 19th, 2013, 07:56 PM   #13
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Re: In what situation frac of limits equals to limit of fr

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 Originally Posted by greg1313 $\lim_{x\to a}\,\frac{f(x)}{g(x)}\,=\,\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$ but what are the conditions on this?
f(x) and g(x) are both continuous at point a and $\lim_{x \to a} g(x)\neq 0$
For some special functions such as $\frac{\sin{x}}{x}$, we believe they are continuous on R.

November 19th, 2013, 09:52 PM   #14
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Re: In what situation frac of limits equals to limit of fr

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 Originally Posted by stainburg Yeah, this Kingkong physicist did the same thing.
<rolled over from the computer desk>.

November 21st, 2013, 07:05 PM   #15
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Re: In what situation frac of limits equals to limit of fr

Quote:
Originally Posted by stainburg
Quote:
 Originally Posted by greg1313 $\lim_{x\to a}\,\frac{f(x)}{g(x)}\,=\,\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$ but what are the conditions on this?
f(x) and g(x) are both continuous at point a and $\lim_{x \to a} g(x)\neq 0$
And are $e^a$ and $10^a$ continuous $\forall\,a\,\in\,\mathbb{R}$? I somewhat understand the need for rigour, but isn't it going too far to say our limit is indeterminate?

November 21st, 2013, 08:54 PM   #16
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Re: In what situation frac of limits equals to limit of fr

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 Originally Posted by greg1313 And are $e^a$ and $10^a$ continuous $\forall\,a\,\in\,\mathbb{R}$
Yes. If we consider e^x and 10^x as the functions that f: R->R, they are continuous at any real point.

Quote:
 Originally Posted by greg1313 but isn't it going too far to say our limit is indeterminate?
I really don't know how to express inf/inf. 'Indeterminate' seems to be a really specified word to those limits where L'Hospital's rule is involved. However, This guy showed that DiracDelta(0)/DiracDelta(0)=1!!!(you know, the Dirac delta function....)

 November 22nd, 2013, 08:25 AM #17 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: In what situation frac of limits equals to limit of fr Dirac delta counts off-topic to the original problem, instead see distribution theory or open a new topic, please. We can approach the inf/inf problem in two ways: 1) in framework of real numbers 2) in framework so-called extended real numbers. In both approach (as I show below), the inf/inf is undefined, that is, it doesn't make any sense. 1) Approching with R Fraction means division. Division (as inverse operator of multiplication) is defined between real numbers. You can not multiply "love" by 2. Why not? Because "love" is not a real number. Inf is not a real number, therefore inf/inf doesn't make sense, it is a syntax error. What is inf then? Inf is a shorthand symbol for an exact statement about limit. Actually you used inf in two places: 'where x tends to' and 'where f(x) tends to'. Let me explain via examples. a) Let f1 be defined on positive real numbers and f1(x) = 1/x. Then f1(x) is a continuous function. 0 doesn't belong to its domain. We can still study the limit at 0. $\lim_{x \to 0} f1(x)= ?$ The limit doesn't exist. f1(x) is divergent at 0. There is no a real number such that f1(x) converges to it. But this is only the half of the truth. This divergence is special: f1(x) is getting bigger than any real number as x tends to 0. This type of divergence is written as $\lim_{x \to 0} f1(x) \to +\infty$ Note that I used arrow instead of equal sign, becasue I wanted to stress that inf is not a number. Unfortunately, some books use an equal sign in this case, pretending as if inf is a number, but you have to know that it's not. b) Let f2 be defined on positive real numbers and f2(x)=sin(1/x). Then f2 is a continuous function. f2 is divergent at zero, but in a different manner than f1. You can not write anything after the arrow. c) Let f3 be defined on R and f3(x) = sin(x). Then f3 is continuous everywhere. Is it continuous at infinity? No. The question is wrong. We can only talk continuity at a point of the domain, but infinity is not a real number. We can still study the limit at infinity of f3(x) and we can show easily it is not convergent, it is divergent, but no special divergence. We can't write anything after the arrow. d) Let f4 defined on R and f4(x) = exp(x). Then f4(x) is divergent in a spacial manner, what we can write concisely as: $\lim_{x \to \infty} f4(x) \to \infty$ (Again, sometimes equal sign used but in the background it refers to the special case of divergence.) 2) Approching with ER By definiton extended reals is a set that consists of every real numbers and two more elements, so-called +inf and -inf. $ER := \{-\infty \cup \mathbb{R} \cup +\infty \}$ As we know, R is not just a pure set, but it is equipped with lots of structure: ring, group, vector space, topological space, metric space etc. Similarly we can try to equip ER with the same structures such that restricting ER on its subset R we get the original structures. In other words, we want to generalize the real numbers. It turn out this generalization can be done for some structures and it can't for others. For example you can easily introduce ordering struture (relatation): $-\infty < \mathbb{R} < +\infty$ There is no contradiction, relation axioms are satisfied. However, you can't introduce addition once you require the usual featues of addition. You can say -inf + 5 = -inf and +inf + 5 = +inf, but what about (+inf) + (-inf)? Let's try to define +inf + -inf as 3.14. In that case you can say: 3.14 = +inf + -inf = (+inf + 5) + -inf = 5 + inf - inf = 5 + 3.14 = 8.14 which is a contradiction. You can check easily that it leads to contradiction whatever you define inf-inf. What about the division? I don't need to mention that you can introduce division for all ER element, because you can't define inf/inf. This imperfection shows clearly that it would be not fair to refer to ER as numbers. However the complex numbers can be called numbers, becuase it has the traditional features that we expect for numbers. ER is just not cool enough. The main reason ER is introduced is its well-behaved topological structure. (Actually ER is homeomorf with the circle.) Let turn back to our functions we studied above and inspect them in this framework. f1: ER+ --> ER. Is divergent f1 at 0? No. It is convergent and limit f1(x) = +inf. Is it continuous at 0? No, because 0 doesn't belong to the domain. But f1 is continuous at +inf, if we define f1(+inf) = 0. Note, now the equal sign is total exact. f2: ER+ --> ER. It is divergent at zero. f2 is convergent and continuous at +inf. f3: ER --> ER. f3 is divergent and non-continuous at inf. f4: ER --> ER. f4 is convergent and continuous at inf. $\lim_{x \to +\infty} e^x= f4(+\inf) = +\inf$ Again, here the equal sign is correct. The limit is an element of ER. __________________________________________________ __________________________________________________ _______________________ Conclusion $\frac{\lim e^x}{\lim 10^x}= 0$ can not be true or false, because it contains a syntax error.
 November 24th, 2013, 01:14 AM #18 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: In what situation frac of limits equals to limit of fr Ok, csak. Thanks for the explanation.

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