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

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

Dave
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    use the periodicity of the sin function: $\sin(a+2\pi)=\sin(a)$ – Exodd Jan 09 '17 at 16:34
  • Thank you, I need understand why $\lim_{x \to 0}\frac{\sin(x)}{x}$ is $1$... – Dave Jan 09 '17 at 16:46
  • @divisor. the expansion of $sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!}...$. Factor out x and, you get - $x[1 - \frac{x^2}{3!} + \frac{x^4}{5!}...]$. In the parantheses, set all terms except the $1$ to zero as $x$ tends to zero. The $x$ is now divided out by the $x$ in the denominator. – Clock Slave Jan 09 '17 at 17:03

4 Answers4

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hint

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

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

hamam_Abdallah
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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.

郑豆浆
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  • I know that $\lim_{x \to 0}\frac{\sin(x)}{x}=1$, why $\lim_{x \to 0}\frac{\sin(x)}{x}$ is $1$ too? – Dave Jan 09 '17 at 16:45
  • $\lim_{x\to 0}\frac{1}{\pi}\left(\frac{\sin\pi x}{\pi x}\right)^{-1}$ – Dave Jan 09 '17 at 16:51
  • Do you mean we already know $\lim_{x \rightarrow 0} \frac{sinx}{x}=1$, why $lim_{x \rightarrow 0} \frac{x}{sinx} = 1$? The limit has its multiplication property and division property. – 郑豆浆 Jan 09 '17 at 16:53
  • Like that, $1=\lim_{x \rightarrow 0} \frac{sinx}{x} \frac{x}{sinx} = \lim_{x \rightarrow 0} \frac{sinx}{x} \lim_{x \rightarrow 0} \frac{x}{sinx} = 1 \cdot \lim_{x \rightarrow 0} \frac{x}{sinx}$. Maybe it is not serious. – 郑豆浆 Jan 09 '17 at 16:58
  • Sorry, $\lim_{x \rightarrow 0} \frac{sinx}{x} \frac{x}{sinx}$ it's my stupidity... – Dave Jan 09 '17 at 17:10
  • I would like to increase more, but I can increase only once :( – Dave Jan 09 '17 at 17:13
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$$\begin{align} \lim_{x\to 0}\frac{x}{\sin\pi(x+2)}&=\lim_{x\to 0}\frac{x}{\sin(\pi x+2\pi)}\\ &=\lim_{x\to 0}\frac{x}{\sin\pi x}\\ &=\lim_{x\to 0}\frac{1}{\pi}\left(\frac{\sin\pi x}{\pi x}\right)^{-1}\\ &=\frac{1}{\pi}1^{-1}=\frac{1}{\pi}. \end{align} $$

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

Dr. Sonnhard Graubner
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