Solving Special Limits not using L'Hopital's Rule

Question

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

Answer 1

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

Answer 2

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

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

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

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

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 $1$.

Answer 3

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

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