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
$$\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right)$$
 A: Hint:
\begin{align}
\lim_{x\to  1^+}\frac{\tan \sqrt{x-1}}{\sqrt{x^2-1}}&=\left(\lim_{x\to1^+}\frac{\sin\sqrt{x-1}}{\sqrt{x-1}}\right)\left(\lim_{x\to1^+}\frac{1}{(\cos\sqrt{x-1})\sqrt{x+1}}\right)\\
\end{align}
A: By the Taylor expansion , as $u \to 0$, one has
$$
\tan u=u+O(u^3)
$$ giving, as $x \to 1^+$,
$$
\tan \sqrt{x-1}=\sqrt{x-1}+O((x-1)^{3/2})
$$ thus
$$
\begin{align}
\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}&=\frac{\sqrt{x-1}}{\sqrt{x-1}\times \sqrt{x+1}}+O\left(\frac{(\sqrt{x-1})^3}{\sqrt{x-1}\times \sqrt{x+1}}\right)\\\\
&=\frac{1}{\sqrt{x+1}}+O\left(\frac{x-1}{\sqrt{x+1}}\right)\\\\
&\to \frac1{\sqrt{2}}.
\end{align}
$$
A: 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! 
A: Another approach is to make a substitution. In this case we can let $x=u^2+1$ with $u>0$ so that $\sqrt{x-1} = \sqrt{u^2} = u$. Then:
$$
\begin{align}
\lim_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right) &= \lim_{u \to 0^+} \left(\frac{\tan u}{\sqrt{(u^2+1)^2-1}}\right)\\
&= \lim_{u \to 0^+} \left(\frac{\tan u}{\sqrt{u^4+2u^2}}\right)\\
&= \lim_{u \to 0^+} \left(\frac{\tan u}{u\sqrt{u^2+2}}\right)\\
&= \lim_{u \to 0^+} \left(\frac{\sin u}u\right)\left(\frac1{\cos u\sqrt{u^2+2}}\right)\\
&= \ldots
\end{align}
$$
This amounts to the same solution as @MarioG's, but the substitution eliminates most of the radicals and puts the $\frac{\sin u}u$ into what you may find a more familiar form.
A: $$
\lim_{x \to 1^+} \left(\frac{\tan\sqrt{x-1} }{\sqrt{x^2-1}}\right)  = \lim_{x \to 1^+} \left(\frac{\tan\sqrt{x-1} }{\sqrt{x-1}\sqrt{x+1}}\right) = \lim_{x \to 1^+} \left(\frac{\sin\sqrt{x-1} }{\sqrt{x-1}} \cdot \frac{1}{\cos\sqrt{x-1}} \cdot \frac{1}{\sqrt{x+1}}\right) =$$ 
$$ \lim_{x \to 1^+} \left(\frac{\sin\sqrt{x-1} }{\sqrt{x-1}} \right) \cdot \lim_{x \to 1^+} \frac{1}{\cos\sqrt{x-1}} \cdot \lim_{x \to 1^+}\frac{1}{\sqrt{x+1}} = 1 \cdot 1 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}
$$
