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
$$\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right)$$

I thought it might be instructive to present a development that goes back to "the basics" and relies on elementary tools only.  To that end, we recall from geometry that the sine function satisfies the inequalities
$$z \cos z \le \sin z\le z \tag 1$$
for $0\le z\le \pi/2$.
Rearranging $(1)$ we immediately see that $z\le \tan z$.  Furthermore, from $(1)$ we see that $\sin^2 z\le z^2$ from which it is straightforward to show that $$\cos z\ge \sqrt{1-z^2}$$ for $0\le z\le 1$.  Therefore, we have
$$z\le \tan z \le \frac{z}{\sqrt{1-z^2}} \tag 2$$
Now, letting $z=\sqrt{x-1}$ in $(2)$ reveals for $1 \le x\le 2$
$$\sqrt{x-1}\le \tan\left(\sqrt{x-1}\right)\le \frac{\sqrt{x-1}}{\sqrt{2-x}} \tag 3$$
Dividing both sides of $(2)$ by $\sqrt{x^2-1}$, we obtain for $1<x\le 2$
$$\frac{1}{\sqrt{x+1}}\le \frac{\tan\left(\sqrt{x-1}\right)}{\sqrt{x^2-1}}\le \frac{1}{\sqrt{x+1}\sqrt{2-x}} \tag 4$$
Applying the Squeeze Theorem to $(4)$ gives the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 1^+}\left(\frac{\tan\left(\sqrt{x-1}\right)}{\sqrt{x^2-1}}\right)=\frac{1}{\sqrt{2}}}$$
and we are done! 
