
#1
Dec2712, 10:23 AM

P: 170

1. The problem statement, all variables and given/known data
Study the continuity of the function defined by: ## \lim n \to \infty \frac{n^xn^{x}}{n^x+n^{x}}## 3. The attempt at a solution I've never seen a limit like this before. The only thing I have thought of is inserting random values of x to see it the limit exists. For instance, in this case, for x=0 I'd have ##\frac{\infty^0\infty^0}{\infty^0+\infty^0}## which means the function doesn't exist. but every other value of x, it's okay. Or am i supposed to solve the limit? (btw, how can I solve a limit for n to infinity??) thank you 



#2
Dec2712, 11:03 AM

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By definition, [itex]n^0 = 1[/itex] for any nonzero [itex]n[/itex], so your fraction reduces to
[tex]\frac{n^0  n^0}{n^0 + n^0} = \frac{1  1}{1 + 1} = \frac{0}{2} = 0[/tex] What does this imly about the limit [tex]\lim_{n \rightarrow \infty} \frac{n^0  n^0}{n^0 + n^0}[/tex]? 



#3
Dec2812, 04:53 AM

P: 170

well, I'd say it means the limit exists and it is = 0 (because I can evaluate it before adding the infinities in the equation).
so the function should continuos on all ##\mathbb{R}##. Or actually, how should i work for x→∞? sorry if i have so many doubts about a simple question, but i have never seen this kind of limits before 



#4
Dec2812, 08:45 AM

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study the continuity of this function 



#5
Dec2912, 01:58 AM

P: 170

ok.
I see that whatever x i choose (except x=0), I always get a 0 in the nominator and an infinity in the denominator, so appartently f(x)=0. But I'm not sure about that.. 



#6
Dec2912, 03:04 AM

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[tex]\frac{n^x  n^{x}}{n^x + n^{x}} = \frac{n  1/n}{n + 1/n} = \frac{1  1/n^2}{1 + 1/n^2}[/tex] What is the limit of this expression as [itex]n \rightarrow \infty[/itex]? 



#7
Dec3012, 04:33 AM

P: 170

and for x=2 as well, because ##\frac{n^2n^{2}}{n^2+n^{2}}##dividing both members by## \frac{1}{n^2} =\frac{1\frac{1}{n^4}}{1+\frac{1}{n^4}}## and so forth ##\forall x \in \mathbb{R}## except x=0, where f(x)=0. Then, can i say it is continuos on all R except between 0 and 1? 



#8
Dec3012, 08:44 AM

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#9
Dec3012, 10:33 AM

P: 170

I've tried several values and found out that if ##x>0 \Rightarrow f(x)=1## ##x=0 \Rightarrow f(x)=0## ##x<0 \Rightarrow \lim f(x)= \frac{\infty}{\infty}## so the function doesn't exist there Also, it is discontinuos in 0. Are my assumptions right? 



#10
Dec3012, 11:50 AM

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#11
Dec3012, 02:56 PM

P: 170

for x=1 I get
## \frac{1\frac{1}{n{^2}}}{1+ \frac{1}{n^{2}}}## ##\frac{1n^2}{1+n^2}= \frac{\infty}{\infty}## If ##\frac{\infty}{\infty}## doesn't mean it's undefined, when can i say the function is discontinuos? 



#12
Dec3012, 03:26 PM

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#13
Dec3012, 03:46 PM

P: 170

right, i didn't notice it, then for x<0 i get f(x)=1.
what can i say if i get ##\frac{\infty}{\infty}##? and when can i say that the function is not continuos, if that's not enough? 



#14
Dec3012, 03:57 PM

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#15
Dec3012, 04:24 PM

P: 170

well, as the right limit is different from the left one, I'd say 0 is a discontinuity point (the only one in R).
fortunately, there are no infinity/infinity cases in this function. but if one of these cases happens with a similar limit, should i just leave it? (if algebric manipulation can't help) thank you very much for your help, anyway :) 



#16
Dec3012, 04:41 PM

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