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3psilonBest ResponseYou've already chosen the best response.0
I knew it had to do with the squeeze theorem that x^2 is throwing me off
 one year ago

LolWolfBest ResponseYou've already chosen the best response.2
We can use a simple identity to solve such: \[\lim_{x\to0}\left(\frac{\sin x}{x^2+4x}\right)=\lim_{x\to0}\left(\frac{\sin x}{x(x+4)}\right)=\lim_{x\to0}\left(\frac{1}{x+4}\right)=\frac{1}{4}\]
 one year ago

LolWolfBest ResponseYou've already chosen the best response.2
(Sorry, I was about to do l'Hospital's and then I saw the factorization at the bottom, @Hero !)
 one year ago

HeroBest ResponseYou've already chosen the best response.1
I suppose my hint was useless
 one year ago
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