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mathslover
 one year ago
Evaluate \(\lim_{x\to 0} \sin^{1} \left( \cfrac{\cos^{1} x + \cos^{1} (x^2) }{\pi} \right) \)
mathslover
 one year ago
Evaluate \(\lim_{x\to 0} \sin^{1} \left( \cfrac{\cos^{1} x + \cos^{1} (x^2) }{\pi} \right) \)

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Hint : \[\lim f(g(x)) = f(\lim g(x))\] if \(f(x)\) is a continuous function

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1I don't know, I'm getting the answer as 0 .. but the book suggests that it doesn't exist :/

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Is it not that when we replace x = 0, we have cos^(1) (0) = pi/2, hence lim sin^(1) (1) = pi/2?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Oh yeah, sorry for that. But, still, how does the limit not exist?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1And here is the reason it has provided: "Limit does not exist as LHL does not exist." \(\lim_{x \to 0^{}} \left( \cfrac{\cos^{1} x + \cos^{1} (x^2)}{\pi} \right) = 1^{+} \)

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Oh okay. Thanks a lot bhaiya! :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6wait a second, maybe book is right... what is the domain of arccos(x) ?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1It is 1 to 1 [1,1]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Ahh right, we're good then. Limit exists and equals pi./2 :)

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1great, thanks again bhaiya

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6\[\large \lim\limits_{x\to 1} ~\arcsin(x) \stackrel{?}{ =}\frac{\pi}{2}\]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Yeah! I got it bhaiya...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6that is a trick question actually, notice that x=1 is a boundary point arcsinx is defined in [1, 1]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Oh, but that will disturb the differentiability of the function, right? How's it going to affect the limit?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6You're probably right, can you give me a convincing argument why the limit exists (recall that both side limits must exist and be same for the overall limit to exist)dw:1434814029787:dw

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1I see what you mean to point here. I'm just confused that we have arcsin(1) here, right? Then why will we take case of \(1^{+}\) here?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Even if we are considering RHL, then that doesn't mean we can simply put 1(+) at the place of 1 there...! right? And the book shows that LHL is arcsin(1(+))

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6we simply don't worry about RHL here because the function itself is not defined for \(x\gt 1\) a sequence can converge happily to a boundary point, no issue with it i guess @zzr0ck3r @eliassaab

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then why don't we do the same for \[\sqrt{x} \] ?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, but here, x is tending to zero. Why will we take case of x >1 ? :/ Sorry if these questions sound stupid but..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6@Catch.me are you suggesting \(\lim\limits_{x\to 0^{}}~\sqrt{x}\) doesn't exist if we think of \(\sqrt{x}\) as a real valued function ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so I can't understand your reasoning!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Okay :) may i knw why do you think \(\lim\limits_{x\to 0^{}}~\sqrt{x}\) doesn't exist ?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1May be because LHL doesn't exist?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1limit exists if \[\lim_{x\to a^+}f(x)=\lim_{x\to a^}f(x)=L\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because the function itself is not defined LHL

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6why do we care about what happens for x<0 where the function itself is not defined

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1if the function is not defined on one side, then a one sided limit exist, not limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is the definition of the limit x goes to 0 not + only or  only

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6that is valid only for the interior limit points

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1just part of the definition there is no proviso that says "if the function does not exist to the right or left, then the one sided limit is the limit"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6x=0 is a boundary point in our specific two cases. so we say the limit exists because we don't care what happens out side the domain

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think wolframalpha isn't concerned with only realvalues

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6i think it makes more sense if we consider sequences definition of limit, let me review it quick and get back

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6meantime here is a nice read http://math.stackexchange.com/questions/637280/limitofsquarerootofxasxapproaches0

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1^ That completely confused me. No one ever taught me about that :/ I was always taught that it doesn't exist

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 can you tell us what is equences definition of limit??

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1The limits of functions in terms of sequences? \[\lim_{x\to c} f(x) = L\]Just that if you consider all sequences \(\{ x_n\}\to c\) then the sequence \(\{f(x_n)\} \to L\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1And the limit of a sequence\[\lim x_n = c\]if there exists an \(M_0\) for all \(\epsilon > 0\) such that for all \(n > M_0\) we have \(x_n c< \epsilon \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6thanks @ParthKohli , that looks neat! I don't seem to remember much of real analysis :/

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1But still, I remain confused :(

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, these definitions look completely nonsensical... until you draw this little picture. I guess you can use these to prove that\[\lim_{x\to 0}\sqrt x =0\]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Actually, I'm talking about the question at the top. Why does the limit not exist?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6just for the record for the main q, the limit exists and equals pi/2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6dw:1434816607981:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6sequences can converge to a boundary point, we don't need to wry about what happens on the other side which is not part of the domain (other side simply doesn't exist as far as this particular problem is concerned) dw:1434816820615:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Wow, that looks really nice.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6that is as much real analysissy i can get, im actually looking for somebody else to explain it more better :)
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