I need to find $\lim _{x \to 0} \cot(3x)\sin(4x)$. However, I am having trouble finding a way to do that. I am a Calculus 1 student and the only ways I know to handle a problem like this are by multiplying by a conjugate, or L'Hospital's Rule. Neither of which seems to work here.

I think I need to identify the correct trig identity for cotangent or sine, and then apply one of the two methods I mentioned, but I can't seem to find any trig identities that seem to allow that.

I have tried replacing $\cot(3x)\sin(4x)$ with $\frac{\cot(3x)}{\csc(4x)}$ and $\frac{\frac{cos(3x)}{sin(3x)}}{\csc(4x)}$ and $\frac{\cos(3x)\sin(4x)}{\sin(3x)}$, but I can't find any that work.

As I understand it, L'Hospital's rule requires that both the numerator and denominator of the limit approach either zero or infinity. I'm not entirely sure what that means, but $(0, 0)$ is a point on both the graph of $\cos(3x)\sin(4x)$ and $\sin(3x)$.

I can do,

$$\lim _{x \to 0}\frac{\frac{d}{dx}\cos(3x)\sin(4x)}{\frac{d}{dx}\sin(3x)} $$

$$=\lim_{x \to 0} \frac{[-3\sin(3x)\cdot\sin(4x)] + [\cos(3x)\cdot4\cos(4x)]}{\cos(3x)}$$

$$=\lim_{x \to 0} \frac{4\cos(3x)\cos(4x)-3\sin(3x)\sin(4x)}{\cos(3x)}$$

$$=\frac{4(1)(1) - 3(0)(0)}{1} = 4$$

But this is wrong.

How can I solve this limit?

  • 1
    $\begingroup$ You are nearly there, $\frac{d}{dx}\sin{(3x)} = \color{red}3\cos{(3x)}$. The other answers below are the usual right way to do it though. $\endgroup$ – user1952500 Apr 7 at 1:04
  • 1
    $\begingroup$ @user1952500 That was all I needed, thank you! $\endgroup$ – LuminousNutria Apr 7 at 1:06

Hint:$cot(3x)sin(4x)=cos(3x){x\over {sin(3x)}}{{sin(4x)}\over x}$

Use the fact that $lim_{x\rightarrow 0}{{sin(x)}\over x}=1$.


$$\cot(3x)\sim_0 \frac{1}{3x}$$ and $$\sin{4x}\sim_0 4x$$ $$...$$ $$\cot{3x}\sin{4x}\sim_0 \frac{4}{3}$$


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