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

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

LolWolf
 2 years ago
Best 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 !)

Hero
 2 years ago
Best ResponseYou've already chosen the best response.1I suppose my hint was useless
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