<s>[INST] What is the value of $\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1}$? $$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = ?$$
I have done these steps to find the answer:


*

*$x^2-1=(x+1)(x-1)$

*$\sqrt{4x-4}=2\sqrt{x-1}$

*$\displaystyle\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = \lim_{x \to 1^+} \frac{\sqrt{ (x+1)(x-1) }+x-1}{ 2\sqrt{x-1} + (x+1)(x-1) }$
So how do I remove what causes the hole function not to become $\frac{0}{0}$ and solve the limit?
 [/INST] Set $x-1=t^2$, which you can because the limit is for $x\to1^+$. Then the limit becomes
$$
\lim_{t\to0^+}\frac{t\sqrt{t^2+2}+t^2}{2t+t^2(t^2+2)}=
\lim_{t\to0^+}\frac{\sqrt{t^2+2}+t}{2+t(t^2+2)}
$$
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