# 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

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

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}

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}

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!

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.

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