Find the limit $\lim_{x\to 1}\frac{\sin{(\pi\sqrt x)}}{\sin{(\pi x)}}$ Find the following limit:
$$\lim_{x\to 1}\frac{\sin{(\pi\sqrt x)}}{\sin{(\pi x)}}$$
My attempt: 
$$t:=x-1,\ x \rightarrow 1 \Rightarrow t\rightarrow 0,\ x=t+1$$
$$\lim_{x\to 1}\frac{\sin{(\pi\sqrt x)}}{\sin{(\pi x)}}=\lim_{t\to 0}\frac{\sin{(\pi\sqrt{t+1})}}{\sin{(\pi(t+1))}}=\lim_{t\to 0}\frac{\frac{\sin{(\pi\sqrt{t+1})}}{\pi(\sqrt{t+1})}\cdot \pi \sqrt{(t+1)}}{\frac{\sin{(\pi(t+1))}}{\pi(t+1)}\cdot \pi(t+1)}=\lim_{t\to 0}\frac{1\cdot \pi \sqrt{(t+1)}}{1\cdot\pi(t+1)}=\frac{\pi\sqrt{(0+1)}}{\pi\sqrt{(0+1)}}=1$$
The soulution should be $\frac{1}{2}$. What am I doing wrong?
 A: Your mistake is here $$\lim _{ t\to 0 } \frac { \sin { \left( \pi \left( t+1 \right)  \right)  }  }{ \pi \left( t+1 \right)  } \neq 1\\ \lim _{ t\to 0 } \frac { \sin { \left( \pi \sqrt { \left( t+1 \right)  }  \right)  }  }{ \pi \sqrt { \left( t+1 \right)  }  } \neq 1$$
A: \begin{eqnarray}
\lim_{x\to1}\dfrac{\sin(\pi\sqrt{x})}{\sin(\pi x)}&=&\lim_{x\to1}\dfrac{\sin(\pi (\sqrt{x}-1))}{\sin(\pi(x-1))}=\lim_{x\to1}\dfrac{\sin(\pi(\sqrt{x}-1))}{\pi(\sqrt{x}-1)}\cdot\lim_{x\to1}\dfrac{\pi(\sqrt{x}-1)}{\pi(x-1)}\cdot\lim_{x\to1}\dfrac{\pi(x-1)}{\sin(\pi(x-1))}\\
&=&\lim_{u\to0}\dfrac{\sin u}{u}\cdot\lim_{v\to1}\dfrac{v-1}{v^2-1}\cdot\lim_{w\to0}\dfrac{w}{\sin w}=1\cdot\lim_{v\to1}\dfrac{1}{v+1}\cdot1=\dfrac12.
\end{eqnarray}
A: As $\sin(\pi-y)=\sin y$
$$\dfrac{\sin(\pi\sqrt x)}{\sin(\pi x)}=\dfrac{\sin\pi(1-\sqrt x)}{\sin\pi(1-x)}$$
Now set $1-\sqrt x=u$
A: A way to work it out could be using Taylor expansions around $x=1$
Then you get:
\begin{equation}
\lim_{x \rightarrow 1}{\dfrac{-\dfrac{1}{2}\pi (x-1) + \dfrac{1}{8}\pi (x-1)^2 + O(x^3)}{-\pi (x-1) + O(x^3)}  }
\end{equation}
Since the terms with the lower power are the ones that tend slower to 0, they are dominating on the expression; therefore:
\begin{equation}
= \lim_{x \rightarrow 1}{\dfrac{-\dfrac{1}{2}\pi (x-1) }{-\pi (x-1)}  } = \dfrac{1}{2}
\end{equation}
