How to calculate $\lim\limits_{x \to 0 } \frac{x}{\sin{\pi(x+2)}}$

Question

I divided the limit by the product of the two limits.

The first limit is found, but how to calculate second: $\lim\limits_{x \to 0 } \frac{x}{\sin{\pi(x+2)}}$

Answer 1

$$\begin{align} \lim_{x\to 0}\frac{x}{\sin\pi(x+2)}&=\lim_{x\to 0}\frac{x}{\sin(\pi x+2\pi)}\\ &=\lim_{x\to 0}\frac{x}{\sin\pi x}\\ &=\lim_{x\to 0}\frac{1}{\pi}\left(\frac{\sin\pi x}{\pi x}\right)^{-1}\\ &=\frac{1}{\pi}1^{-1}=\frac{1}{\pi}. \end{align} $$

Answer 2

hint

$$\sin(\pi(x+2))=\sin(x\pi+2\pi)$$ $$=\sin(x\pi )$$

$$\lim_{X\to 0}\frac{X}{\sin(X)}=1$$

Answer 3

with the help of the rules of L'Hospital we obtain $$\lim_{x \to 0}\frac{x}{\sin(\pi(x+2))}=\lim_{x \to 0}\frac{1}{\pi\cos(\pi(x+2))}=...$$

Answer 4

Assume we already know $\lim_{x \rightarrow 0 \frac{x}{sinx}}=1$.

$\lim_{x \rightarrow 0 \frac{x}{sin{\pi (x+2)}}} = \lim_{x \rightarrow 0} \frac{x}{sin{\pi x}} = \lim_{\pi x \rightarrow 0} \frac{\pi x}{sin{\pi x}} \cdot \frac{1}{\pi} = \frac{1}{\pi} \lim_{u \rightarrow 0} \frac{u}{sinu} = \frac{1}{\pi}$

As for $\lim_{x \rightarrow 0 \frac{x}{sinx}}=1$, it needs another proof.