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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

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HINT

We have that

$$\frac{\sin(4x)}{x^2+x}=\frac{\sin(4x)}{4x}\frac{4x}{x^2+x}$$

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I've written some possible approaches.

FIRST

\begin{align*} \lim_{x\rightarrow 0}\frac{\sin(4x)}{x^{2} + x} = \lim_{x\rightarrow 0}\frac{\sin(4x)}{x(x+1)} = \lim_{x\rightarrow 0}\frac{4\sin(4x)}{4x}\times\lim_{x\rightarrow 0}\frac{1}{x+1} = \lim_{y\rightarrow 0}\frac{4\sin(y)}{y} = 4 \end{align*}

SECOND \begin{align*} \lim_{x\rightarrow 0}\frac{\sin(4x)}{x^{2} + x} & = \lim_{x\rightarrow 0}\frac{\sin(4x) - \sin(0)}{4x - 0}\times\lim_{x\rightarrow 0}\frac{4}{1+x} = 4\times\lim_{y\rightarrow 0}\frac{\sin(y) - \sin(0)}{y - 0} = 4\cos(0) =4 \end{align*}

THIRD

Since the Taylor series of the sine function is given by

\begin{align*} \sin(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \ldots \end{align*}

We have $\sin(4x) \sim 4x + O(x^{3})$ when $x$ is near to zero. Hence we obtain: \begin{align*} \lim_{x\rightarrow 0}\frac{\sin(4x)}{x^{2}+x} = \lim_{x\rightarrow 0}\frac{4x}{x^{2}+x} = \lim_{x\rightarrow 0}\frac{4}{x+1} = 4 \end{align*}

FOURTH

\begin{align*} \lim_{x\rightarrow 0}\frac{\sin(4x)}{x^{2}+x} = \lim_{x\rightarrow 0}\frac{2\sin(2x)\cos(2x)}{x(x+1)} = 4\times\lim_{x\rightarrow 0}\frac{\sin(2x)}{2x}\times\lim_{x\rightarrow 0}\frac{\cos(2x)}{x+1} = 4\times 1\times 1 = 4 \end{align*}

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

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A very simple, yet working approach is the following:

Taking a limit of sin(x) with x going towards 0 can be reduced to the limit of x towards 0.

This means that:

$$\lim_{x\rightarrow 0} \dfrac{sin(4x)}{x^2 + x}=\lim_{x\rightarrow 0} \dfrac{4x}{x^2 + x} = \infty$$

The reason for that is that $sin(x) = x$ for $x \approx0$. There are multiple proofs of this. One is done by @gimusi.

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