$ \lim_{x \to 0}{\frac{\sin( \pi \cos x)}{x \sin x} }$ I have to evaluate the following limit
$$ \lim_{x \to  0}{\frac{\sin( \pi \cos x)}{x \sin x} }$$
My solution is:
$$ \lim_{x \to  0}{\frac{\sin( \pi \cos x)}{x \sin x} }=\lim_{x \to  0}{\frac{\pi \cos x}{x \cdot  x} }=+\infty$$
But the correct result is $\frac{\pi}{2}$. I can't understand where I'm making mistakes.
 A: The error, as noticed in the comments, is that $\pi\cos x$ is not close to $0$.
Simpler to use asymptotic analysis: near $0$, we have $\:\cos x=1-\dfrac{x^2}2+o(x^2)$, and $\sin u\sim_0 u$, so
\begin{align}
\frac{\sin(\pi\cos x)}{x\sin x}&=\frac{\sin\Bigl(\pi-\cfrac{\pi x^2}{2}+o(x^2)\Bigr)}{x \sin x} =\frac{\sin\Bigl(\cfrac{\pi x^2}{2}+o(x^2)\Bigr)}{x\sin x} \\
&\sim_0\frac{\cfrac{\pi \not{\!x^2}}{2}+o(\not{\!x^2})}{\not{\!x^2}}=\frac{\pi}{2}+o(1)
\end{align}
A: There are many ways in which you can calculate this. For example: L'Hospital's rule applied once will get you to
$$\underset{x\to 0}{\text{lim}}-\frac{\pi  (x \sin ) (\cos  (\pi  (x \cos )))}{x (\cos  x)+\sin  x}$$
(I leave the derivatives to you, as a warm up).
Now by product rule you can split the fraction as a product:
$$-\pi  \left(\underset{x\to 0}{\text{lim}}\cos  (\pi  (\cos  x))\right) \underset{x\to 0}{\text{lim}}\frac{\sin  x}{x (\cos  x)+\sin  x}$$
Easily observing that the brackets limit is $-1$ which then makes $-\pi$ to $\pi$.
Now you can can multiply and divide by $x$ to get
$$\pi \lim_{x\to 0} \frac{\color{red}{\sin(x)}}{x\cos(x) + \sin(x)}\cdot \frac{x}{\color{red}{x}}$$
The red zone goes to $1$. Collect $x$ over and above:
$$\pi \lim_{x\to 0} \frac{x}{x\left(\cos(x) + \frac{\color{red}{\sin(x)}}{\color{red}{x}}\right)}$$
Again a red zone that goes to $1$.
$x$ can be simplified.
The cosine of $x\to 0$ goes to $1$ leading to
$$\frac{\pi}{1+1} = \frac{\pi}{2}$$
A: Only using

*

*well know limits,

*the identity $\sin\theta=\sin(\pi-\theta)$

*the fact that $\pi(1-\cos{x})\to0$ when $x\to0$:
\begin{align}
  &\lim_{x\to0}{\frac{\sin(\pi\cos{x})}{x\sin{x}}}=\\
  &=\lim_{x\to0}{\frac{\sin(\pi-\pi\cos{x})}{x\sin{x}}}=\\
  &=\lim_{x\to0}{\frac{\sin[\pi(1-\cos{x})]}{\pi(1-\cos{x})}\cdot\frac{\pi(1-\cos{x})}{x\sin{x}}}=\\
  &=\lim_{x\to0}{\frac{\sin[\pi(1-\cos{x})]}{\pi(1-\cos{x})}\cdot\pi\cdot\frac{1-\cos{x}}{x^2}\cdot\frac{x}{\sin{x}}}=\\
  &=1\cdot\pi\cdot\frac{1}{2}\cdot1=\frac{\pi}{2}\\
\end{align}
A: Use the identity $\>\pi\cos x=\pi\bigl(1-2\sin^2{x\over2}\bigr)$ to obtain
$${\sin(\pi\cos x)\over x\sin x}={\sin\bigl(2\pi\sin^2{x\over2}\bigr)\over x\cdot2\sin{x\over2}\cos{x\over2}}={\pi\over2}\cdot{\sin\bigl(2\pi\sin^2{x\over2}\bigr)\over2\pi\sin^2{x\over2}}\cdot{\tan{x\over2}\over{x\over2}}\ .$$
On the right hand side the two last factors tend to $1$ when $x\to0$. It follows that
$$\lim_{x\to0}{\sin(\pi\cos x)\over x\sin x}={\pi\over2}\ .$$
A: By some elementary inequalities, near $x=0$ we have
$$
\cos(x)=1-x^2/2+O(x^4)$$In particular, near zero we have
$$
\frac{\sin(\pi\cos(x))}{x\sin(x)} = \frac{\sin(\pi(1-x^2/2)+ O(x^4))}{x\sin(x)}
$$Use the sine-reflection and angle-addition:
$$
 \frac{\sin(\pi(1-x^2/2)+O(x^4))}{x\sin(x)} = \frac{\sin(\pi/2 \cdot x^2+O(x^4))}{x\sin(x)}
$$
$$
= \frac{\sin(\pi/2\cdot x^2)\cos(O(x^4))}{x\sin(x)}+ \frac{\cos(\pi/2 \cdot x^2)\sin(O(x^4))}{x\sin(x)}
$$Using $\lim_{\theta\to 0}\frac{\sin(\theta)}{\theta}=1$, the second term vanishes in the limit, and $\cos(O(x^4))\to 1$. So we have
$$
\lim_{x\to 0}\frac{\sin(\pi/2 \cdot x^2)\cdot 1}{x\sin(x)} = \lim_{x\to 0}\frac{\sin(\pi/2 \cdot x^2)}{x^2}\cdot \frac{x}{\sin(x)} =\frac{\pi}{2}\cdot 1 =\frac{\pi}{2}
$$
A: $$L=\lim_{x \to 0}{\frac{\sin( \pi \cos x)}{x \sin x} }$$
$$L=\lim_{x \to 0}{\frac{\sin( \pi \cos x)}{x^2} }$$
By L'Hôpital's rule:
$$L=-\lim_{x \to 0}{\frac{\cos( \pi \cos x)\pi \sin x}{2x} }$$
$$L=-\lim_{x \to 0}{\frac{\cos( \pi \cos x)\pi }{2} }$$
$$\implies L=\dfrac {\pi}2$$
