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Bomull
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Do you know l'Hôpital's Rule?

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

abb0t
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Heroism attempt #1 ^^ What about the denominator?

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

SithsAndGiggles
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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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