## 3psilon 2 years ago Lim x->0 sinx/x^2+4x why is it 1/4?!

1. Hero

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

2. 3psilon

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

3. LolWolf

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

4. 3psilon

AHhhh I see it!

5. LolWolf

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

6. 3psilon

Thanks @LolWolf

7. Hero

I suppose my hint was useless