# Solving Special Limits not using L'Hopital's Rule [closed]

How to find the following limit$$\lim_{x\to 0}\frac{\tan(5x)}{\sin\left(\frac{x}{3}\right)}$$

## closed as off-topic by kingW3, Simply Beautiful Art, Trevor Gunn, Leucippus, Bobson DugnuttJul 4 '17 at 22:38

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• $\dfrac{\tan 5x}{\sin\frac x3} = \dfrac{\tan 5x}{x}\cdot\dfrac{x}{\sin\frac x3}$ – peterwhy Jul 4 '17 at 21:41
• Piggy backing @peterwhy : $$\frac{\tan(5x)}{\sin(x/3)}=\frac{\sin(5x)}{5x}\frac{x/3}{\sin(x/3)}\frac{15}{\cos(5x)}$$ – Simply Beautiful Art Jul 4 '17 at 21:43
• In the future, please format your math using MathJax and to ask well-received questions, please include context, such as what you've tried, where you saw this problem, etc.. We're not here to do your homework for you. – Simply Beautiful Art Jul 4 '17 at 21:45
• So please understand why I am downvoting (for now.) I'm downvoting because this question is low quality, shows no efforts on your behalf, and not even any research effort. Note however, that you can edit your question to improve it up to site standards (upon which I'd happily remove my downvote.) – Simply Beautiful Art Jul 4 '17 at 21:49

If you're willing to do a bit of trig and a little induction, it's possible to compute this limit without knowing that ${\sin x\over x}\to1$ as $x\to0$. Note first that

$$\sin((n+1)x)=\sin(nx)\cos(x)+\sin x\cos(nx)$$

so if we know that

$$\lim_{x\to0}{\sin(nx)\over\sin x}=n$$

(which is obviously true for the base case $n=1$), then by induction we have

$$\lim_{x\to0}{\sin((n+1)x)\over\sin x}=n\lim_{x\to0}\cos x+\lim_{x\to0}\cos(nx)=n+1$$

It follows that

$$\lim_{x\to0}{\tan(5x)\over\sin(x/3)}=\lim_{x\to0}{\tan(15x)\over\sin x}=\lim_{x\to0}{\sin(15x)\over\sin x}\cdot{1\over\cos(15x)}=15\cdot1=15$$

Using equivalents: $\;\tan u\sim_0 u \sim_0\sin u$, hence $$\frac{\tan 5x}{\sin\dfrac x3}\sim_0\frac{5\not x}{\dfrac {\not x}3}=15.$$

$$\tan (5x)=\frac {\sin (5x)}{\cos (5x)}$$

$$\sin (5x)=\sin (x/3+14x/3)$$ $$=\sin (x/3)\cos (14x/3)+\cos (x/3)\sin (14x/3)$$

$$\sin (14x/3)=\sin (x/3+13x/3)$$ $$=\sin (x/3)\cos (13x/3)+\cos (x/3)\sin (13x/3)$$

$$\sin (2x/3)=2\sin (x/3)\cos (x/3)$$

the limit is then $$L=1+1+1+1+1+1+1+1+1+1+1+1+1+2=15$$ We used the fact that all $\cos$ go to $1$.