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:


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


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



We have that



I've written some possible approaches.


\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*}


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*}


\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*}


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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.