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Any hints on where I could start with limx→0x²+sin(x²)xsin(x)+cos(x)−1
 one year ago
 one year ago
Any hints on where I could start with limx→0x²+sin(x²)xsin(x)+cos(x)−1
 one year ago
 one year ago

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

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

BomullBest ResponseYou've already chosen the best response.0
Not 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
 one year ago

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

abb0tBest ResponseYou've already chosen the best response.0
You can separate them: \[\lim_{x \rightarrow 0} \frac{ x^2 }{ xsin(x)+\cos(x)1 }+ \lim_{x \rightarrow 0}\frac{ \sin(x^2) }{xsin(x)+\cos(x)1 }\]
 one year ago

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

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

SithsAndGigglesBest ResponseYou've already chosen the best response.2
As 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...
 one year ago

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

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

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

SithsAndGigglesBest ResponseYou've already chosen the best response.2
Alright, 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
 one year ago
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