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Bomull
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0}\frac{ x²+\sin (x²) }{ x \sin (x) + \cos (x) 1 }\]

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.0Do you know l'Hôpital's Rule?

Bomull
 one year ago
Best ResponseYou've already chosen the best response.0Not yet, as that only comes in the second part of the course  but this is a review exercise for the first part exam so there should be a way to compute this without l'hôpital's rule

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.0Anyone capable of doing this thing w/o l'Hôpital is a hero, imo.

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0You can separate them: \[\lim_{x \rightarrow 0} \frac{ x^2 }{ xsin(x)+\cos(x)1 }+ \lim_{x \rightarrow 0}\frac{ \sin(x^2) }{xsin(x)+\cos(x)1 }\]

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.0Heroism attempt #1 ^^ What about the denominator?

Bomull
 one year ago
Best ResponseYou've already chosen the best response.0Some hoodoo with trig identities?

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2As suggested by @abb0t, for the first limit, you can write \[\lim_{x\to 0}\frac{x^2}{x\sin x+\cos x1}\] \[\large \lim_{x\to 0}\frac{x}{\sin x+\frac{\cos x1}{x}}\] Now, I'm not sure about the validity of this next step, but I'm pretty sure because the function is continuous for all nonzero x, you can use the fact that \[\frac{\cos x1}{x}\to0 \text{ as } x\to0,\text{ leaving you with }\\ \lim_{x\to0}\frac{x}{\sin x}=1\] And that takes care of the first limit. As for \[\lim_{x\to0}\frac{\sin(x^2)}{x\sin x+\cos x1},\] I'll need a bit more time...

Bomull
 one year ago
Best ResponseYou've already chosen the best response.0Whoops I accidentally closed this question. Anyways, I'm waiting to see if @SithsAndGiggles can solve the other half!

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2Actually, it looks like a mistake was made somewhere along the line... The first limit should come out to be 2...

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2Unfortunately, WolframAlpha isn't being very accommodating with the whole "not using L'Hopital's rule" thing.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2Alright, I'm thinking of using the following now (so everything I did above should be disregarded): \[\text{Near }x=0, \text{ we have }\\ \sin x\approx x\\ \sin^2x\approx (x)^2=x^2\\ \sin(x^2)\approx x^2\] \[\lim_{x\to0}\frac{x^2+\sin^2x}{x\sin x+\cos x1}\\ \lim_{x\to0}\frac{x^2+x^2}{x\sin x+\cos x1}\\ 2\lim_{x\to0}\frac{x^2}{x\sin x+\cos x1}\\ 2\lim_{x\to0}\frac{x^2}{x\sin x+\cos x1}\cdot\frac{\cos x+1}{\cos x+1}\\ 2\lim_{x\to0}\frac{x^2(\cos x+1)}{x\sin x(\cos x+1)+(\cos^2 x1)}\\ 2\lim_{x\to0}\frac{x^2(\cos x+1)}{x\sin x(\cos x+1)\sin^2x}\\ 2\lim_{x\to0}\frac{x^2(\cos x+1)}{\sin x\left(x(\cos x+1)\sin x\right)}\] Now, using the fact that \[\frac{x}{\sin x}\to1 \text{ as } x\to0,\\ 2\lim_{x\to0}\frac{x(\cos x+1)}{x(\cos x+1)\sin x}\] Since \[\sin x\approx x,\\ 2\lim_{x\to0}\frac{x(\cos x+1)}{x(\cos x+1) x}\\ 2\lim_{x\to0}\frac{\cos x+1}{\cos x+1 1}\\ 2\lim_{x\to0}\frac{\cos x+1}{\cos x}\\ 2\cdot\frac{1+1}{1}\\ 4\] And as a check: http://www.wolframalpha.com/input/?i=limit+of+%28x%5E2%2Bsin%28x%5E2%29%29%2F%28xsinx%2Bcosx1%29+as+x+approaches+0
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