Question: I've been stuck on this problem on my online math system centre, and I've tried everything. I have to send in work for the professor to mark, so I cannot graph it or use plot points. Additionally, I CANNOT use L'Hospital's rule, as it is not covered yet in course material. I apologize in advance, I'm new on StackExchange, but here is the problem:

$$\lim\_{x\to0}\dfrac{\sin4x}{x^2+x}$$

So far, I took the derivative of the equation and got

$$\frac{4\cos4x-2x^{-1}\sin4x-x^{-2}\sin(4x)}{x^{2}+x}$$

But since the limit is still zero, it will still be indeterminate. Since the denominator is $x^{2}+x$, it will always be some variant of that to the nth power. So it's become clear to me I cannot solve this with derivatives. I thought about using $$\lim\_{h\to0} \frac{f(x+h)-f(x)}h$$ but that didn't work out either. 

Can someone help me here? Regards, Joe

Answer: 
**HINT**

We have that

$$\frac{\sin(4x)}{x^2+x}=\frac{\sin(4x)}{4x}\frac{4x}{x^2+x}$$
Write your term as $$\frac{\sin(4x)}{4x}\cdot \frac{4}{x+1}$$