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?

Your mistake is here $$\lim _{ t\to 0 } \frac { \sin { \left( \pi \left( t+1 \right)  \right)  }  }{ \pi \left( t+1 \right)  } \neq 1\\ \lim _{ t\to 0 } \frac { \sin { \left( \pi \sqrt { \left( t+1 \right)  }  \right)  }  }{ \pi \sqrt { \left( t+1 \right)  }  } \neq 1$$
