How to compute $\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used? Thanks
$$\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right)$$

$$
\lim_{x \to 1^+} \left(\frac{\tan\sqrt{x-1} }{\sqrt{x^2-1}}\right)  = \lim_{x \to 1^+} \left(\frac{\tan\sqrt{x-1} }{\sqrt{x-1}\sqrt{x+1}}\right) = \lim_{x \to 1^+} \left(\frac{\sin\sqrt{x-1} }{\sqrt{x-1}} \cdot \frac{1}{\cos\sqrt{x-1}} \cdot \frac{1}{\sqrt{x+1}}\right) =$$ 
$$ \lim_{x \to 1^+} \left(\frac{\sin\sqrt{x-1} }{\sqrt{x-1}} \right) \cdot \lim_{x \to 1^+} \frac{1}{\cos\sqrt{x-1}} \cdot \lim_{x \to 1^+}\frac{1}{\sqrt{x+1}} = 1 \cdot 1 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}
$$
