Question

Lim x->0 sinx/x^2+4x why is it 1/4?!

Answer 1

Hint: sin(x)/x --> 1

Answer 2

I knew it had to do with the squeeze theorem that x^2 is throwing me off

Answer 3

We can use a simple identity to solve such:
\[\lim_{x\to0}\left(\frac{\sin x}{x^2+4x}\right)=\lim_{x\to0}\left(\frac{\sin x}{x(x+4)}\right)=\lim_{x\to0}\left(\frac{1}{x+4}\right)=\frac{1}{4}\]

Answer 4

AHhhh I see it!

Answer 5

(Sorry, I was about to do l'Hospital's and then I saw the factorization at the bottom, @Hero !)

Answer 6

Thanks @LolWolf

Answer 7

I suppose my hint was useless